Problem 1.1
The force, F , of the wind blowing against a building is given by F CD V 2 A 2, where V is
the wind speed, the density of the air, A the cross-sectional area of the building, and C D is
a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Solution 1.1
F
CD V 2
A
2
or
CD
2F
V 2A
, where
F  MLT 2 ,  ML 3, V  LT 1, A  L2
Thus,
CD 
MLT 2
ML 3 LT
1 2
M 0 L0T 0
L2
Hence, C D is dimensionless.
Problem 1.2
The Mach number is a dimensionless ratio of the velocity of an object in a fluid to the speed
of sound in the fluid. For an airplane flying at velocity V in air at absolute temperature T ,
the Mach number M a is,
Ma
V
,
kRT
where k is a dimensionless constant and R is the specific gas constant for air. Show that M a
is dimensionless.
Solution 1.2
We denote the dimension of temperature by
Then
L
T
Ma
1
or
Ma
1.
FL
M
L
T
ML
T 2F
L2
T2
and use Newton’s second law to get F
ML
T2
.
Problem 1.3
Verify the dimensions, in both the FLT and the MLT systems, of the following quantities,
which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Volume, (b) acceleration, (c) mass, (d) moment of inertia (area), and (e) work.
Solution 1.3
a) volume  L3
b) acceleration time rate of change of velocity 
LT 1
 LT 2
T
c) mass  M
or with F  MLT 2
mass  FL 1T 2
d) moment of inertia area
e) work
second moment of area  L2
force distance  FL
or with F  MLT 2
work  ML2T 2
L2  L4
Problem 1.4
Verify the dimensions, in both the FLT and the MLT systems, of the following quantities,
which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Angular velocity, (b) energy, (c) moment of inertia (area), (d) power, and (e) pressure.
Solution 1.4
angular displacement
T 1
time
b) energy ~ capacity of body to do work
single work force distance ➔ energy  FL
a) angular velocity
or with F  MLT 2 ➔ energy  MLT 2
c) moment of inertia area
L  ML2T 2
second moment of area  L2
FL
 FLT 1  MLT 2
T
L2  L4
L T 1  ML2T 3
d) power
rate of doing work 
e) pressure
force F
 2  FL 2  MLT 2 L 2  ML 1T 2
area L
Problem 1.5
Verify the dimensions, in both the FLT system and the MLT system, of the following
quantities, which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Frequency, (b) stress, (c) strain, (d) torque, and (e) work.
Solution 1.5
cycles
T 1
time
force
F
b) stress =
 2  FL 2
area
L
a) frequency =
Since F  MLT 2 ,
stress 
MLT 2
2
 ML 1T 2
L
change in length L
c) strain =
  L0 dimensionless
length
L
d) torque = force distance  FL  MLT 2 L  ML2 T 2
e) work = force distance  FL  MLT 2 L  ML2 T 2
Problem 1.6
If u is velocity, x is length, and t is time, what are the dimensions (in the MLT system) of
(a) u / t , (b) 2 u / x t , and (c) ( u / t )dx?
Solution 1.6
a)
u LT 1

 LT 2
t
T
2
u
LT 1
b)

T 2
x t ( L )(T )
c)
LT
u
x
t
T
1
(L)  L2T 2
Problem 1.7
Verify the dimensions, in both the FLT system and the MLT system, of the following
quantities, which appear in Table B.1 Physical Properties of Water (BG/EE Units).
(a) Acceleration, (b) stress, (c) moment of a force, (d) volume, and (e) work.
Solution 1.7
L
velocity
 2  LT 2
time
T
force
F
b) stress
 2  FL 2
area
L
Since F  MLT 2 ,
a) acceleration
stress 
MLT 2
2
 ML 1T 2
L
c) moment of a force
force distance  FL  MLT 2 L  ML2T 2
d) volume (length)3  L3
e) work
force distance  FL  MLT 2 L  ML2T 2
Problem 1.8
If p is pressure, V is velocity, and
is fluid density, what are the dimensions (in the MLT
system) of (a) p / , (b) pV , and (c) p / V 2?
Solution 1.8
a)
p

FL 2
ML
3
MLT 2 L 2
ML
b) pV  ML 1T 2
c)
p
V
2

3
ML
3
 L2T 2
LT 1 ML 3  M 2 L 3T 3
ML 1T 2
ML 3 LT
ML 1T 2
1 2
 M 0 L0T 0 dimensionless
Problem 1.9
If P is force and x is length, what are the dimensions (in the FLT system) of (a) dP / dx ,
(b) d 3 P / dx3, and (c) P dx ?
Solution 1.9
dP F
a)
  FL 1
dx L
b)
c)
d 3P
dx
3

F
3
L
 FL 3
Pdx  FL
Problem 1.10
If V is velocity, is length, and
2
is a fluid property (the kinematic viscosity) having
1
dimensions of L T , which of the following combinations are dimensionless: (a) V
(b) V , (c) V 2 , and (d) V ?
Solution 1.10
a) V
b)
V
 LT 1 L L2T 1  L4T 2 not dimensionless

LT 1 L
LT
1
c) V 2  LT 1
2
d)
V

2
 L0T 0 dimensionless
L2T 1  L4T 3 not dimensionless
LT 1
2
L LT
1
 L 2 not dimensionless
,
Problem 1.11
The momentum flux is given by the product mV , where m is mass flow rate and V is velocity.
If mass flow rate is given in units of mass per unit time, show that the momentum flux can
be expressed in units of force.
Solution 1.11
M
T
mV
where
1
gc
L
T
M
L
T2
F T2
ML
F
FT2
comes from Newton’s Second Law.
ML
Problem 1.12
An equation for the frictional pressure loss p (inches H2O) in a circular duct of inside diameter d in. and length L ft for air flowing with velocity V ft/min is
p 0.027
L
d 1.22
V
Vo
1.82
,
where V0 is a reference velocity equal to 1000 ft/min. Find the units of the “constant” 0.027.
Solution 1.12
Solving for the constant gives
pL
0.027
L
D1.22
V
Vo
1.82
.
The units give
0.027
in. H2 O
ft
in.1.22
0.027
ft
min
ft
min
in. H2O in.1.22
ft
1.82
Problem 1.13
The volume rate of flow, Q, through a pipe containing a slowly moving liquid is given by
the equation
R4 p
Q
8
where R is the pipe radius, p the pressure drop along the pipe, is a fluid property called
viscosity FL 2T , and is the length of pipe. What are the dimensions of the constant
Would you classify this equation as a general homogeneous equation? Explain.
Solution 1.13
3
LT
1

L3T 1 
L4
8
8
FL 2
FL 2T
L
L3T 1
The constant is
is dimensionless.
8
Yes. This is a general homogeneous equation
because it is valid in any consistent units system.
/8?
Problem 1.14
Show that each term in the following equation has units of lb/ft3 . Consider u as velocity, y
as length, x as length, p as pressure, and as absolute viscosity.
0
2
p
x
u
y2
.
Solution 1.14
lb
ft 2
ft
p
x
lb
p
x
or
ft3
,
and
u
lb sec
ft
sec
y2
ft 2
ft 2
2
2
or
u
lb
y2
ft3
.
Problem 1.15
The pressure difference, p, across a partial blockage in an artery (called a stenosis) is approximated by the equation
V
Kv
D
p
Ku
A0
1
A1
2
V2
where V is the blood velocity,
is the blood viscosity ( FL 2T ),
is the blood density
ML 3 , D is the artery diameter, A0 is the area of the unobstructed artery, and A1 is the area
of the stenosis. Determine the dimensions of the constants K and K u . Would this equation
be valid in any system of units?
Solution 1.15
V
Kv
D
p
Ku
A0
1
A1
2
V2
L2
FT L 1
FL  Kv 2
L T L
Ku
FL 2  Kv FL 2
K u FL 2
2
L2
2
1
FT 2 1
L L3
L
T
2
K and K u are dimensionless because each term in the equation must have the same
dimensions. Yes, The equation would be valid in any consistent system of units.
Problem 1.16
Assume that the speed of sound, c , in a fluid depends on an elastic modulus, E , with dimensions FL 2, and the fluid density, , in the form c
E
a
b
. If this is to be a dimen-
sionally homogeneous equation, what are the values for a and b ? Is your result consistent
E
with the standard formula for the speed of sound? (See the equation c
.)
Solution 1.16
Substituting c
LT 1
LT 1
a
FL 2
FL 2
E
FL 4T 2
b
FL 4T 2 into the equation provided yields:
F a b L 2 a 4 bT 2 b
Dimensional homogeneity requires that the exponent of each dimension on both sides
of the equal sign be the same.
F: 0 = a + b
L: 1 = −2a − 4b
T: −1 = 2b
Therefore:
T: −1 = 2b ➔ b = −1/2
F: a = −b ➔ a = 1/2
L:
1 = −2a−4b = −2(1/2) −4(−1/2)= 1
a
1
; b
2
1
2
Yes, this is consistent with the standard formula
for the speed of sound.
Problem 1.17
A formula to estimate the volume rate of flow, Q, flowing over a dam of length, B , is given
by the equation
Q
3.09 BH 3/2
where H is the depth of the water above the top of the dam (called the head). This formula
gives Q in ft3 /s when B and H are in feet. Is the constant, 3.09, dimensionless? Would this
equation be valid if units other than feet and seconds were used?
Solution 1.17
Q
3
3.09 BH 2
3
L3T 1  3.09 L L 2
5
L3T 1  3.09 L 2
Since each term in the equation must have the same dimensions the constant
1
3.09 must have dimensions of L2 T 1 and is therefore not dimensionless. No.
Since the constant has dimensions its value will change with a change in units.
No.
Problem 1.18
A commercial advertisement shows a pearl falling in a bottle of shampoo. If the diameter D
of the pearl is quite small and the shampoo is sufficiently viscous, the drag on the pearl is
given by Stokes’s law,
3 VD,
where V is the speed of the pearl and is the fluid viscosity. Show that the term on the right
side of Stokes’s law has units of force.
Solution 1.18
3 VD
M
LT
L
L
L M 2
T
T
M
FT 2 L
ML T 2
1
gc
where
1
gc
FT2
ML
comes from Newton’s Second Law.
F
Problem 1.20
Express the following quantities in SI units: (a) 10.2 in. min, (b) 4.81 slugs, (c) 3.02 lb ,
(d) 73.1 ft s2 , and (e) 0.0234 lb s ft 2 .
Solution 1.20
a) 10.2
in.
min
10.2
b) 4.81slugs
c) 3.02 lb
ft
d) 73.1 2
s
e) 0.0234
2.540 10 2
4.81slugs 1.459 10
N
lb
3.02 lb
4.448
ft
73.1 2
s
m
2
3.048 10 1 s
ft
s2
lb s
ft
in.
min
2
0.0234
lb s
ft
2
m
in.
kg
slug
1 min
60 s
4.32 10 3
70.2 kg
13.4 N
m
22.3 2
s
N s
2
4.788 10 m
lb s
ft 2
1.12
N s
m2
m
s
4.32
mm
s
Problem 1.21
Express the following quantities in BG units: (a) 14.2 km , (b) 8.14 N m3, (c) 1.61kg m3 ,
(d) 0.0320 N m s, and (e) 5.67 mm hr .
Solution 1.21
a) 14.2 km
b) 8.14
c) 1.61
N
m3
kg
m
3
d) 0.0320
e) 5.67
14.2 103 m 3.281
8.14
1.61
N m
s
mm
hr
N
m3
kg
m
3
ft
m
4.66 10 4 ft
lb
3
6.366 10 3 ft
N
m3
5.18 10 2
slugs
3
1.940 10 3 ft
kg
m3
3.12 10 3
0.0320
5.67 10 3
N m
s
m
hr
ft lb
7.376 10 1 s
N m
s
3.281
ft
m
1 hr
3600 s
lb
ft3
slugs
ft3
2.36 10 2
5.17 10 6
ft
s
ft lb
s
Problem 1.22
Express the following quantities in SI units: (a) 160 acres , (b) 15 gallons (U.S.), (c) 240 miles ,
(d) 79.1 hp, and (e) 60.3 °F.
Solution 1.22
ft 2
acre
9.290 10 2
liters
gallon
10 3
a) 160 acre
160 acre 4.356 10 4
b) 15 gallons
15 gallons 3.785
c) 240 mi
d) 79.1 hp
240 mi 5280
ft
mi
ft lb
79.1 hp 550 s
hp
3.048 10 1
1.356
m
ft
J
ft lb
e) Relationship between units of temperature:
5 o
K o C 273
F 32 273
9
5
60.3o F 32 273 289 K
9
m3
liter
m2
ft
2
6.47 105 m 2
56.8 10 2 m3
3.86 105 m
5.90 10 4
J 1W
s 1 Js
5.90 10 4 W
Problem 1.23
Water flows from a large drainage pipe at a rate of 1200 gal min. What is this volume rate
of flow in (a) m3 s, (b) liters min, and (c) ft3 s ?
Solution 1.23
a) flowrate
1200
gal
min
m3
6.309 10 5 s
gal
min
7.57 10 2
m3
s
b) Since1 liter 10 3 m3
flowrate
c) flowrate
7.57 10 2
m3
s
2 m
s
3
103 liters
m
3
60 s
min
ft3
3.531 10 s3
m
s
4540
2.67
ft3
s
liters
min
Problem 1.24
The universal gas constant R0 is equal to 49,700 ft 2 / s2 °R , or 8310 m 2 / s2 K . Show
that these two magnitudes are equal.
Solution 1.24
R0
8310 m 2
s2 K
3.281ft
1m
2
5/9 K
1°R
49,700
ft 2
sec 2 °R
Problem 1.25
Dimensionless combinations of quantities (commonly called dimensionless parameters)
play an important role in fluid mechanics. Make up five possible dimensionless parameters
by using combinations of some of the quantities listed in Table B.1 Physical Properties of
Water (BG/EE Units).
Solution 1.25
Some possible examples:
2
LT
T
acceleration time

 L0T 0
1
velocity
LT
frequency time  T 1 T  T 0
2
2
LT 1
velocity

 L0T 0
2
length acceleration
L LT
F T
F T
force time


 F 0 L0T 0
1
2
1
1
momentum
MLT
FT L
LT
ML 3 LT 1 L
density velocity length

 M 0 L0T 0
1
1
dynamic viscosity
ML T
Problem 1.26
An important dimensionless parameter in certain types of fluid flow problems is the Froude
number defined as V g , where V is velocity, g is the acceleration of gravity, and is
32.2 ft s2 , and
length. Determine the value of the Froude number for V 10 ft s , g
2 ft . Recalculate the Froude number using SI units for V , g, and . Explain the significance of the results of these calculations.
Solution 1.26
In BG units,
ft
s
ft
32.2 2 2 ft
s
10
V
g
1.25
In SI units:
V
10
g
9.81
2 ft
ft
m
0.3048
s
ft
m
3.05
m
s
s2
0.3048
m
ft
0.610 m
Thus,
V
g
3.05
m
9.81 2
s
m
s
1.25
0.610 m
The value of a dimensionless parameter is independent of the unit system.
Problem 1.28
A tank contains 500 kg of a liquid whose specific gravity is 2. Determine the volume of the
liquid in the tank.
Solution 1.28
m
V SG H 2OV
Thus,
V=
m
SG H2O
500 kg
kg
2 999 3
m
0.250 m3
Problem 1.29
A stick of butter at 35°F measures 1.25in. 1.25in. 4.65in. and weighs 4 ounces. Find its
specific weight.
Solution 1.29
4 oz
W
V
1.25 in.
2
1 lb
16 oz
1 ft
4.65 in.
12 in.
3
59.5
lb
ft3
Problem 1.30
Clouds can weigh thousands of pounds due to their liquid water content. Often this content
is measured in grams per cubic meter g/m3 . Assume that a cumulus cloud occupies a
volume of 1 cubic kilometer, and its liquid water content is 0.2 g/m3. (a) What is the volume
of this cloud in cubic miles? (b) How much does the water in the cloud weigh in pounds?
Solution 1.30
a) Volume
b) W
1 km
3
3
9
10 m
3.281 ft
1m
1m
5,280 ft
3
3
0.240 mi3
Volume
= g = 0.2
W
1,000 m
1 km
1.962 10 3
g
m
3
N
m
2
m
1 N s2
1 kg m
1 kg
1,000 g
9.81
109 m3
2.248 10 1
s
2
lb
N
1.962 10 3
4.4 105 lb
N
m3
Problem 1.31
A tank of oil has a mass of 25 slugs. (a) Determine its weight in pounds and in Newtons at
the Earth’s surface. (b) What would be its mass (in slugs) and its weight (in pounds) if
located on the moon’s surface where the gravitational attraction is approximately one-sixth
that at the Earth’s surface?
Solution 1.31
a) weight
mass g
ft
25 slugs 32.2 2
s
25 slugs
b) mass
weight
14.59 kg
1 slug
1 lb s2
1 slug ft
805 lb
m
9.81 2
s
1 N s2
1 kg m
3580 N
25 slugs mass does not depend on gravitational attraction
805 lb
6
134 lb
Problem 1.32
A certain object weighs 300 N at the Earth’s surface. Determine the mass of the object (in
kilograms) and its weight (in newtons) when located on a planet with an acceleration of
gravity equal to 4.0 ft s2 .
Solution 1.32
weight
mass
g
For g
300 N
m
9.81 2
s
30.6 kg
ft
4.0 2 ,
s
weight = 30.6 kg
ft
4.0 2
s
0.3048
m
ft
1 N s2
1 kg m
37.3 N
Problem 1.33
The density of a certain type of jet fuel is 775 kg/m3. Determine its specific gravity and
specific weight.
Solution 1.33
kg
m3
kg
1000 3
m
0.775
m
9.81 3
s
7.60
775
SG
o
H 2O @ 4 C
g
775
kg
m
3
kN
m3
Problem 1.34
At 4°C a mixture of automobile antifreeze (50% water and 50% ethylene glycol by volume)
has a density of 1064 kg/m3 . If the water density is 1000 kg/m3, find the density of the
ethylene glycol.
Solution 1.34
meg
S
mw
mixture 4°C
meg
4 °C
mw
water 4°C
mw
mw
4 °C
where mw is the mass of the pure water in volume V at 4°C .
Then
S
0.5
eg
0.5
w
0.5
eg
w
w
.
w
The problem statement gives
kg
m3
kg
1000 3
m
1064
S
mixture
w
➔ 1.064
0.5
1.064
eg
w
w
➔
eg
1.064
1
0.5
kg
1130 3
m
w
eg
1000
kg
m3
1.064
1
0.5
DISCUSSION If the mixture were at some temperature T , then for equal volumes of
mixture and 4°C water,
S
meg
mw
mw 4°C
T
0.5
eg
w T
w 4°C
Problem 1.35
A hydrometer is used to measure the specific gravity of liquids. For a certain liquid, a
hydrometer reading indicates a specific gravity of 1.15.What is the liquid’s density and
specific weight? Express your answer in SI units.
Solution 1.35
SG
H2O @ 4o C
1.15
1000
kg
m3
1.15 1000
g
1150
kg
m3
kg
m3
1150
m
9.81 2
s
kg
m3
1 N s2
1 kg m
11.3
kN
m3
Problem 1.36
An open, rigid-walled, cylindrical tank contains 4 ft3of water at 40 °F . Over a 24-hour
period of time the water temperature varies from 40 to 90 °F. Make use of the data in
Appendix B to determine how much the volume of water will change. For a tank diameter
of 2 ft , would the corresponding change in water depth be very noticeable? Explain.
Solution 1.36
mass of water
Amount ot mass is not a function of temperature.
40o
40o
m
90o
90o
From Table B.1 Physical Properties of Water (BG/EE Units)
H2O @40 o F
H2O @90 o F
1.940
1.931
slugs
ft3
slugs
ft3
Therefore,
slugs
ft3
slugs
1.931 3
ft
4 ft3 1.940
V 90 o
4.0186 ft3
4.0186 4.000
Thus, the increase in volume is:
For a tank diameter of 2 ft:
h
0.0186 ft3
A
2
0.0186 ft3
5.92 10 3 ft
0.0710 in.
2 ft
4
This small change in depth would not be very noticeable. No.
Note: A slightly different value for h will be obtained if specific weight of water is used
rather than its density. This is due to the fact that there is some uncertainty in the
fourth significant figure of these two values, and the solution is sensitive to this uncertainty.
Problem 1.38
A mountain climber’s oxygen tank contains 1 lb of oxygen when he begins his trip at sea level
where the acceleration of gravity is 32.174 ft/s 2. What is the weight of the oxygen in the tank
when he reaches the top of Mt. Everest where the acceleration of gravity is 32.082 ft/s 2?
Assume that no oxygen has been removed from the tank; it will be used on the descent
portion of the climb.
Solution 1.38
W
mg
Let
sl
denote sea level and
MtE
denote the top of Mt. Everest
Thus,
Wsl
1 lb
msl g sl
and
WMtE
mMtE gMtE
However msl
msl
Wsl
gsl
mMtE so that since m
mMtE
W
g
WMtE
gMtE
or
WMtE
g
Wsl MtE
gsl
ft
32.082 2
s
1 lb
ft
32.174 2
s
0.9971 lb
Problem 1.39
The information on a can of pop indicates that the can contains 355 mL . The mass of a full
can of pop is 0.369 kg, while an empty can weighs 0.153 N . Determine the specific weight,
density, and specific gravity of the pop and compare your results with the corresponding
values for water at 20 °C . Express your results in SI units.
Solution 1.39
weight of fluid
volume of fluid
total weight
mass g
m
0.369 kg 9.81 2
s
1 N s2
1 kg m
1 m3
1,000 L
355 10 6 m3
3.62 N
weight of can 0.153 N
volume of fluid
355 10 3 L
Therefore,
3.62 N 0.153 N
9770
355 10 6 m3
N
m3
m
9.81 2
s
9770
g
SG
H 2O @ 4 oC
996
N
m3
N s2 1 kg m
m
4
kg
m3
kg
1000 3
m
1N s
2
996
kg
m3
0.996
For water at 20 °C (see Table B.2 Physical Properties of Water [SI Units])
H2O
9789
N
m
3
H2O
998.2
kg
m3
SG
0.9982
A comparison of these values for water with those for the pop shows that the specific
weight, density, and specific gravity of the pop are all slightly lower than the corresponding
values for water.
Problem 1.40
The variation in the density of water, , with temperature, T , in the range
20 °C T 50 °C , is given in the following table.
Density (kg/m3) 998.2 997.1 995.7 994.1 992.2 990.2 988.1
20
25
30
35
40
45
50
Temperature ( C)
Use these data to determine an empirical equation of the form
c1
c2T
c3T 2 which
can be used to predict the density over the range indicated. Compare the predicted values
with the data given. What is the density of water at 42.1 C ?
Solution 1.40
Fit the data to a second order polynomial using a standard curve-fitting program such as
found in Excel. Thus,
1001 0.0533T 0.0041T2
As shown in the table below, (predicted) from Eq.(1) is in good agreement with (given).
T
(oC)
-predict
(kg/m^3)
-data
(kg/m^3)
20
25
30
35
40
45
50
998.3
997.1
995.7
994.1
992.3
990.3
988.1
998.2
997.1
995.7
994.1
992.2
990.2
988.1
At T
42.1 °C :
1001 0.0533 42.1 °C
0.0041 42.1 °C
2
991.5
kg
m3
Problem 1.41
If 1 cup of cream having a density of 1005 kg/m3 is turned into 3 cups of whipped cream,
determine the specific gravity and specific weight of the whipped cream.
Solution 1.41
Mass of cream, m
1005
kg
volume.
cup , were
m3
Noting that the mass is the same in liquid and in “whipped” form,
cream
kg
m3
3 cups
3 cups
mwhipped
whipped
cream
whipped
cream
SG
1005
335
H 2O @ 4 oC
whipped
cream
whipped
cream
g
kg
m3
kg
m3
335
cup
1005
3
kg
m3
335
kg
m3
0.335
kg
m
3
2
m
s
1 N s2
1 kg m
3,290
N
m3
Problem 1.42
With the exception of the 410 bore, the gauge of a shotgun barrel indicates the number of
round lead balls, each having the bore diameter of the barrel, that together weigh 1 lb. For
1
example, a shotgun is called a 12 - gauge shotgun if a -lb lead ball fits the bore of the
12
barrel. Find the diameter of a 12 - gauge shotgun in inches and millimeters. Lead has a
specific weight of 0.411 lb/in3 .
Solution 1.42
weight
ball
For a sphere:
4 R3
3
1
lb
12
lb
0.411 3
in.
0.20276 in.3
R
3
R
3
3
4
3 0.20276 in.3
D
D
0.3644 in.
4
0.729 in.
0.729 in.
25.4 mm
in.
D 18.5 mm
Problem 1.44
A regulation basketball is initially flat and is then inflated to a pressure of approximately
24 lb/in 2 absolute. Consider the air temperature to be constant at 70 °F . Find the mass of
air required to inflate the basketball. The basketball’s inside radius is 4.67 in.
Solution 1.44
m
p
RT
lb
in.2
4
3
1716
ft lb
slug o R
24
m
p 4
R3
RT 3
4.67 in.
0.000938slug
3
1 ft
12 in.
460 70 o R
Problem 1.45
Nitrogen is compressed to a density of 4 kg/m3 under an absolute pressure of 400 kPa .
Determine the temperature in degrees Celsius.
Solution 1.45
T
TC
p
R
TK
N
m2
J
296.8
kg K
400 103
kg
4 3
m
1 Nsm
1J
273 337 K 273 64 0C
337 K
Problem 1.46
The temperature and pressure at the surface of Mars during a Martian spring day were determined to be 50 C and 900 Pa, respectively. (a) Determine the density of the Martian
atmosphere for these conditions if the gas constant for the Martian atmosphere is assumed
to be equivalent to that of carbon dioxide. (b) Compare the answer from part (a) with the
density of the Earth’s atmosphere during a spring day when the temperature is 18 C and
the pressure 101.6 kPa (abs).
Solution 1.46
a)
b)
Mars
900
p
RT
Earth
0
50 C 273 K
1J
101.6 103
J
286.9
kg K
kg
m3
kg
1.22 3
m
0.0214
1 Nsm
J
188.9
kg K
p
RT
N
m2
N
m2
1 Nsm
1J
0.0214
Thus,
Mars
Earth
0.0175 1.75%
1.22
0
C 273 K
kg
m3
kg
m3
Problem 1.47
A closed tank having a volume of 2 ft3 is filled with 0.30 lb of a gas. A pressure gage attached
to the tank reads 12 psi when the gas temperature is 80 F . There is some question as to
whether the gas in the tank is oxygen or helium. Which do you think it is? Explain how you
arrived at your answer.
Solution 1.47
weight
g volume
Density of gas in tank
Since
with T
0.30 lb
ft
32.2 2 2 ft3
s
4.66 10 3
slugs
ft3
p
with p 12 14.7 psia (atmospheric pressure assumed to be 14.7 psia) and
RT
80 F 460 R it follows that
lb
in2
26.7 2 144 2
in
ft
R 540 R
7.12 slugs
R ft3
(1)
From Table 1.7 Approximate Physical Properties of Some Common Gases at Standard
Atmospheric Pressure (BG Units)
R 1.554 103
ft lb
ft lb
for oxygen … and … R 1.242 10 4
for helium.
slug °R
slug °R
Thus, from Eq.(1) if the gas is oxygen
7.12
slugs
3
1.554 10
ft
3
4.58 10 3
slugs
5.73 10 4
slugs
ft3
and for helium
7.12
1.242 10
slugs
4
ft
3
ft3
A comparison of these values with the actual density of the gas in the tank indicates that
the gas must be oxygen.
Problem 1.48
Assume that the air volume in a small automobile tire is constant and equal to the volume
between two concentric cylinders 13 cm high with diameters of 33 cm and 52 cm. The air in
the tire is initially at 25°C and 202 kPa . Immediately after air is pumped into the tire, the
temperature is 30°C and the pressure is 303 kPa . What mass of air was added to the tire?
What would be the air pressure after the air has cooled to a temperature of 0 °C?
Solution 1.48
The mass of air added to the tire is the difference of the final mass of
air m f and the initial mass mi . Assuming air is an ideal gas,
mf
p
RT
mi
pf
p
RT
f
i
h
h = 13 cm
d1 = z r1 = 33 cm
d2 = z r2 = 52 cm
pi
.
Ti
R Tf
r1
r2
Now
r22
r12
26 cm
h
2
16.5 cm
2
1m
13cm
100 cm
3
0.0165 m3 .
0.0165 m3
mf
mi
287.0
mf
mi
N m
kg K
303kPa
273 30 K
202 kPa
273 25 K
N
m2
kPa
1000
0.0185kg
Now consider the cooling process. The initial state will be 30°C and 303kPa. The final state
will be 0°C and p f . Applying the ideal gas law to both states gives
p
RT
Since
p
RT
i
f
i,
pf
pi
Tf
Ti
.
f
303 kPa
273 0
273 30
pf
273 kPa
Problem 1.49
A compressed air tank contains 5 kg of air at a temperature of 80 C . A gage on the tank
reads 300 kPa . Determine the volume of the tank.
Solution 1.49
mass
volume
p
RT
volume
300 101 103
J
286.9
kg K
5 kg
kg
3.96 3
m
1 Nsm
1J
1.26 m3
N
m3
80 C 273 K
3.96
kg
m3
Problem 1.50
A rigid tank contains air at pressure of 90 psia and a temperature of 60 °F. By how much
will the pressure increase as the temperature is increased to 110 °F?
Solution 1.50
p
RT
For a rigid closed tank, the air mass and volume are constant so
the equation above (with R constant)
p1
T1
p2
T2
(1)
where p1
90 psia , T1
60 F 460
and
110 F 460
570 R .
T2
520 R,
From Eq.(1)
p2
T2
p1
T1
570 °R
520 °R
90 psia
98.7 psia
constant . Thus, from
Problem 1.51
The density of oxygen contained in a tank is 2.0 kg/m3 when the temperature is 25 °C .
Determine the gage pressure of the gas if the atmospheric pressure is 97 kPa .
Solution 1.51
p
RT
2.0
kg
m
3
259.8
J
kg K
25 °C 273 K
155 kPa (abs)
p(gage) Pabs Patm 155 kPa 97 kPa 58 kPa
Problem 1.52
The helium-filled blimp shown in the figure below is used at various athletic events. Determine the number of pounds of helium within it if its volume is 68,000 ft 3 and the temperature and pressure are 80 °F and 14.2 psia, respectively.
Solution 1.52
W
68000 ft 3 and
V where V
p
g
RT
g
Thus,
lb
in.2
144
in.2
ft 2
14.2
1.242 10 4
9.82 10 3
ft lb
slug °R
slug
2
ft s
2
80 460 °R
l lb
slug ft
s2
9.82 10 3
Hence,
W
9.82 10 3
lb
ft
3
68000 ft3
ft
32.2 2
s
668 lb
lb
ft3
Problem 1.53
Develop a computer program for calculating the density of an ideal gas when the gas pressure in Pascals (abs), the temperature in degrees Celsius, and the gas constant in J/kg K are
specified. Plot the density of helium as a function of temperature from 0 °C to 200 °C and
pressures of 50, 100, 150, and 200 kPa (abs).
Solution 1.53
For an ideal gas
p
RT
so that
p
RT
where p is absolute pressure, R the gas constant, and T is absolute temperature. Thus, if the
temperature is in °C then
T
°C
273.15 K
A spreadsheet (EXCEL) program for calculating
follows.
This program calculates the density of an ideal gas when the absolute
pressure in Pascal, the temperature in degrees C, and the gas
constant in J/kg·K are specified. To use, replace current values with
desired values of temperature, pressure, and gas constant.
A
B
C
D
Pressure
Temperature
Gas constant
Destiny
Pa
°C
J/kg·K
kg/m3
1.01E+05
15
286.9
1.22
Formula:
=A10/((B10+273.15)*C10)
Example: Calculate
for p
200 kPa, temperature 20 °C, and R
287
J
.
kg K
This program calculates the density of an ideal gas when the
absolute pressure in Pascal, the temperature in degrees Celsius,
and the gas constant in J/kg·K are specified. To use, replace
current values with desired values of temperature, pressure, and
gas constant.
A
Pressure
Pa
2.00E+05
B
Temperature
°C
20
C
Gas constant
J/kg·K
287
D
Destiny
kg/m3
2.38
The density of helium is plotted in the graph below.
Density of Helium
0.40
0.35
0.30
0.25
p
0.20
kg/m3
p = 200 kPa(abs)
0.15
= 150
0.10
= 100
0.05
= 50
0.00
0
50
100
T, °C
150
200
Problem 1.55
For flowing water, what is the magnitude of the velocity gradient needed to produce a shear
N
stress of 1.0 2 ?
m
Solution 1.55
du
where
dy
Thus,
du
dy
1.0
1.12 10 3
N
m2
1.12 10 3
N s
m2
N s
893
m2
1
s
and
1.0
N
m2
Problem 1.56
Make use of the data in Appendix B to determine the dynamic viscosity of glycerin at 85 F .
Express your answer in both SI and BG units.
Solution 1.56
5
TC
TF 32
9
5
85 F 32
9
29.4 C
From the figure in Appendix B:
glycerin at 85 F 29.4 C
0.6
N s
m
2
2.089 10 2
0.6
lb s
ft 2
N s
m2
N s
m2
SI units
1.3 10 2
lb s
ft 2
BG units
Problem 1.57
One type of capillary-tube viscometer is shown in the figure below. For this device the liquid to be tested is drawn into the
tube to a level above the top etched line. The time is then obtained for the liquid to drain to the bottom etched line. The
kinematic viscosity, , in m 2 /s is then obtained from the equation
KR 4t where K is a constant, R is the radius of the capillary tube in mm , and t is the drain time in seconds. When
is used as a calibration fluid in a particular
glycerin at 20
viscometer, the drain time is 1430 s. When a liquid having a
density of 970 kg/m3 is tested in the same viscometer the drain
time is 900 s . What is the dynamic viscosity of this liquid?
Glass
strengthening
bridge
Etched lines
Capillary
tube
Solution 1.57
KR 4t
For glycerin @ 20 C
m2
1.19 10 3
s
KR
For unknown liquid with t
4
7
8.32 10
m2
s2
900 s
8.32 10
By definition:
KR 4 1430 s
2
= 970
kg
m3
7
m2
s2
900 s
7.49 10 4
m2
s
7.49 10
0.727
0.727
4
m2
s2
kg 1 N s 2
m s 1 kg m
N s
m2
Problem 1.58
The viscosity of a soft drink was determined by using a
capillary tube viscometer shown in the figure. For this device
the kinematic viscosity, , is directly proportional to the time, t,
that it takes for a given amount of liquid to flow through a
small capillary tube. That is,
Kt. The following data were
obtained from regular pop and diet pop. The corresponding
measured specific gravities are also given. Based on these data,
by what percent is the absolute viscosity, μ, of regular pop
greater than that of diet pop?
Regular pop
Diet pop
t(s)
377.8
300.3
SG
1.044
1.003
Glass
strengthening
bridge
Etched lines
Capillary
tube
Solution 1.58
reg
% greater
diet
100
diet
By definition
Given
reg
1
100
diet
, and
SG
H 2O @4 C.
kt :
% greater
reg
diet
% greater 31.0 %
1
100
t SG reg
t SG diet
1
100
377.8 s 1.044
300.3 s 1.003
1
100
Problem 1.59
The viscosity of a certain fluid is 5 10 4 poise. Determine its viscosity in both SI and BG
units.
Solution 1.59
From Appendix E, 1 poise 10 1
Thus,
5 10 4 poise
N s
m2
N s
10 1 2
m
1 poise
.
5 10 5
N s
m2
.
From Table 1.4 Conversion Factors from SI Units to BG and EE Units (end paper)
5 10 5
N s
m
2
2.089 10 2
N s
1 2
m
lb s
ft 2
10.4 10 7
lb s
ft 2
Problem 1.60
The kinematic viscosity and specific gravity of a liquid are 3.5 10
4
m2
and 0.79 , respecs
tively. What is the dynamic viscosity of the liquid in SI units?
Solution 1.60
SG
H 2O @ 4 C
3.5 10 4
m2
s
0.79 103
kg
m3
0.277
kg 1 N s 2
m s 1 kg m
0.277
N s
m2
Problem 1.61
A liquid has a specific weight of 59 lb/ft 3 and a dynamic viscosity of 2.75 lb s/ft 2 . Determine
its kinematic viscosity.
Solution 1.61
By definition:
, and
g
2.75
g
lb s
ft 2
59
,
32.2
lb
ft3
ft
s2
1.50
ft 2
s
Problem 1.62
The kinematic viscosity of oxygen at 20 C and a pressure of 150kPa (abs) is 0.104 stokes .
Determine the dynamic viscosity of oxygen at this temperature and pressure.
Solution 1.62
p
RT
150 103
259.8
J
kg K
1N m
1J
0.104 stokes
0.104
cm 2
s
cm 2
0.104
s
2.05 10
5
1m
100 cm
kg 1 N s2
m s 1 kg m
N
m2
1.97
20 C 273 K
2
1.97
kg
m3
2.05 10 5
N s
m2
kg
m3
Problem 1.63
Fluids for which the shearing stress, , is not linearly related to the rate of shearing strain, ,
are designated as non-Newtonian fluids. Such fluids are commonplace and can exhibit
unusual behavior. Some experimental data obtained for a particular non-Newtonian fluid
at 80 F are shown below.
(lb/ft 2 ) 0
2.11 7.82 18.5 31.7
(s 1 )
50
0
100
150
200
Plot these data and fit a second-order polynomial to the data using a suitable graphing program. What is the apparent viscosity of this fluid when the rate of shearing strain is 70 s 1?
Is this apparent viscosity larger or smaller than that for water at the same temperature?
Solution 1.63
Shearing
stress,
lb/sq ft
0
50
100
150
200
0
2.11
7.82
18.5
31.7
Shearing stress, lb/sq ft
Rate of
shearing
strain, 1/s
τ = 0.0008γ⋅ 2 + 0.0035γ⋅
40
30
20
10
0
0
50
100
150
200
Rate of shearing strain, 1/s
From the graph
0.0008 2
0.0035 where
is the shearing stress in
lb
ft 2
and
is the rate
of shearing strain in s-1. Fitting a second-order polynomial to the data yields:
d
2 0.0008
0.0035
apparent
d
At
70 s 1
apparent
2 0.0008
lb s2
ft
2
70 s 1
0.0035
lb s
ft
2
0.116
lb s
ft 2
From Table B.1 Physical Properties of Water (BG/EE Units)
5 lb s
. Water is a Newtonian fluid so this value is independent of
H2O @ 80 F 1.791 10
ft 2
. Thus, the viscosity of the non-Newtonian fluid when sheared at a rate of 70 s-1 is much
larger than the viscosity of water at 80 F .
Problem 1.64
Water flows near a flat surface and some measurements of the water velocity, u , parallel to
the surface, at different heights, y, above the surface are obtained. At the surface y 0.
After an analysis of the data, the lab technician reports that the velocity distribution in the
range 0 y 0.1 ft is given by the equation
u 0.81 9.2 y 4.1 103 y3
with u in ft/s when y is in ft . (a) Do you think that this equation would be valid in any system of units? Explain. (b) Do you think this equation is correct? Explain.
Solution 1.64
(a)
u
0.81 9.2 y 4.1 103 y3
LT 1  0.81
9.2 L
4.1 103
L3
Each term in the equation must have the same dimensions. Thus, the constant 0.81
must have dimensions of LT 1, 9.2 dimensions of T 1, and 4.1 103 dimensions of
L 2T 1. Since the constants in the equation have dimensions their values will change
with a change in units. No.
(b) Equation cannot be correct since at y
0 u
0.81
would violate the “no-slip” condition. Not correct.
ft
, a non-zero value which
s
Problem 1.65
Calculate the Reynolds numbers for the flow of water and for air through a 4-mm-diameter
tube, if the mean velocity is 3m/s and the temperature is 30 C in both cases. Assume the
air is at standard atmospheric pressure.
Solution 1.65
For water at 30 C (from Table B.2 Physical Properties of Water [SI Units]):
kg
995.7 3
m
N s
7.975 10 4
m2
kg
m
995.7 3 3
0.004 m
VD
1 N s2
s
m
Re
15000
1
kg
m
4 N s
7.975 10
m2
For air at 30 C (from the Physical Properties of Air at Standard Atmospheric Pressure
[SI Units]):
1.165
kg
m3
1.86 10 5
Re
VD
N s
m2
1.165
kg
m3
3
1.86 10
m
s
5
0.004 m
N s
m2
1 N s2
1 kg m
752
Problem 1.66
SAE 30 oil at 30 F flows through a 2-in.-diameter pipe with a mean velocity of 5 ft/s.
Determine the value of the Reynolds number.
Solution 1.66
slugs
1.77
ft3
8.0 10 3
Re
VD
lb s
ft 2
1.77
slugs
ft3
ft 2
ft
s 12
lb s
8.0 10 3
ft 2
5
1 lb s2
1 slug ft
184
Problem 1.67
For air at standard atmospheric pressure, the values of the constants that appear in the
CT 3/2
are C 1.458 10 6 kg/ m s K1/2 and S 110.4 K. Use
Sutherland equation
T S
these values to predict the viscosity of air at 10 C and 90 C and compare with values given
in the table Physical Properties of Air at Standard Atmospheric Pressure (SI Units).
Solution 1.67
kg
T 3/2
1/2
m s K
T 110.4 K
3
CT 2
1.458 10 6
T S
For T
10 C
T=10 273.15 K
283.15 K,
3
kg
283.15
K
2
m s K1/2
283.15 K 110.4 K
1.458 10 6
1.765 10
5
2
kg
3/2 1/2 1 1 N s
K
m s
1 kg m
1.76 10 5
From the table,
For T
1.765 10 5
90 C 90 C 273.15 363.15 K,
1.458 10 6 363.15
3/2
363.15 110.4
From the table,
2.13 10 5
N s
2.14 10 5
N s
m2
m2
N s
m2
N s
m2
.
Problem 1.68
Use the values of viscosity of air given in the table of Physical Properties of Air at Standard
Atmospheric Pressure (SI Units) at temperatures of 0, 20, 40, 60, 80, and 100 C to deterCT 3/2
.
mine the constants C and S, which appear in the Sutherland equation
T S
Compare your results with the values given in Problem 1.67.
T 3/2
T 3/2
1
S
(Hint: Rewrite the equation in the form
and plot
versus T . From the
T
C
C
slope and intercept of this curve, C and S can be obtained.)
Solution 1.68
3
T2
3/2
CT
1
S
T
(1)
can be written in the form
T S
C
C
Entering the specified temperatures and the corresponding viscosities from the table yields
the 1st and 3rd columns in the following table. The 2nd column is a conversion of the temperatures to an absolute scale and the 4th column contains the values of the LHS of Eq. 1 for
these temperatures and viscosities.
Equation
T
C
T K
0
20
40
60
80
100
273.15
293.15
313.15
333.15
353.15
373.15
Plotting
T 3/2
T 3/2
N s
m2
1.71×10-5
1.82×10-5
1.87×10-5
1.97×10-5
2.07×10-5
2.17×10-5
K 3/2
kg
m s
2.640×108
2.758×108
2.963×108
3.087×108
3.206×108
3.322×108
versus. T yields:
3.5 × 108
3.3 × 108
T 3/2
μ
3.1 × 108
2.9 × 108
2.7 × 108
2.5 × 108
240
273
293
313
333
353
373
400
T (K)
A polynomial of order one, which is a straight line, would be a reasonable fit for the data.
Using Excel to determine the constants for a fit of the data to a straight line given by
y bx a
T 3/2
1
S
where x T , y
, b
, and a
.
C
C
yields:
y 6.969 105 x 7.441 107.
Therefore:
1
C
b 6.969 105
C 1.43 10 6
And
kg
m s K3/2
S
a 7.441 107
C
S 107 K
These values for C and S are in good agreement with values given in Problem 1.67.
Problem 1.69
The viscosity of a fluid plays a very important role in determining how a fluid flows. The
value of the viscosity depends not only on the specific fluid but also on the fluid temperature. Some experiments show that when a liquid, under the action of a constant driving
pressure, is forced with a low velocity, V , through a small horizontal tube, the velocity is
given by the equation V K . In this equation K is a constant for a given tube and pressure, and is the dynamic viscosity. For a particular liquid of interest, the viscosity is given
by Andrade’s equation
B
T
De
with D 5 10 7 lb s ft 2 and B 4000 R . By what percentage will the velocity increase as
the liquid temperature is increased from 40 F to 100 F ? Assume all other factors remain
constant.
Solution 1.69
K
V40
(1)
40
V100
K
(2)
100
V100 V40
V40
% increase in V
V100
V40
100
1
100
and from Eq. (1) and (2)
K
% increase in V
100
K
1
100
100
1
100
40
40
From Andrade’s equation
40
5 10 7 e
4000
40 F 460
and
7
100
5 10 e
4000
100 F 460
Thus, from Eq. (3)
% increase in V
5 10
5
7
4000
e 500
4000
7 560
10 e
1
100 136 %
(3)
Problem 1.70
Use the value of the viscosity of water given in Table B.2 Physical Properties of Water (SI
Units) at temperatures of 0, 20, 40, 60, 80, and 100
to determine the constants D and B
De B /T . Calculate the value of the viscosity at 50 C
which appear in Andrade’s equation
and compare with the value given in the table above. (Hint: Rewrite the equation in the
1
form ln
B
ln D and plot ln versus 1 T . From the slope and intercept of this curve,
T
B and D can be obtained. If a nonlinear curve-fitting program is available, the constants
can be obtained directly from Andrade’s equation
equation.)
De B /T without rewriting the
Solution 1.70
De B /T can be written in the form
Equation
ln
B
1
T
(1)
ln D
and with data from the table in the problem:
1
N s
T C T K
T (K)
m2
0
273.15
3.661×10−3
1.787×10−3
20
293.15
3.411×10−3
1.002×10−3
−3
40
313.15
3.193×10
6.529×10−4
60
333.15
3.002×10−3
4.665×10−4
80
353.15
2.832×10−3
3.547×10−4
100
373.15
2.680×10−3
2.818×10−4
A plot of ln
versus
ln
−6.327
−6.906
−7.334
−7.670
−7.944
−8.174
1
is shown below:
T
–9.0
–8.0
ln μ
–7.0
–6.0
2.0 × 10−3
3.0 × 10−3
4.0 × 10−3 T (K)
Although there appears to be a slight curvature to the data in the semi-log plot, it also appears to be reasonably well approximated by a straight line as would be expected for data
that follows a n exponential law. Using an exponential law (y
the same as fitting a straight-line on a semi-log plot), yields:
a 1.767 10 6
D
aebx) fit in Excel, (which is
N s
m2
and
B
b 1.870 103 K
so that
1.767 10
6
1870
e T
At 50 C (323.15 K),
1.767
From the table in the problem,
1870
6 323.15
10 e
5.76 10 4
5.468 10 4
N s
m2
N s
m2
Problem 1.71
7.1 10 5 lb s ft 2 at 40 F and
1.9 10 5 lb s ft 2 at 150 F .
For a certain liquid
Make use of these data to determine the constants D and B , which appear in Andrade’s
De B /T . What would be the viscosity at 80 F ?
equation
Solution 1.71
De B /T
At T
40 F 459.67
499.67 R,
at T
150 F 459.67
609.67 R,
7.1 10 5
lb s
, and
ft 2
lb s
1.9 10 5
.
ft 2
Take the logarithm of both sides of the equation
ln
B
1
T
DeD/T to yield
(1)
ln D
Substitute above values of
and T into Eq. (1) to give
ln 7.1 10 5
B
1
499.67
ln D
(2)
ln 1.9 10 5
B
1
609.67
ln D
(3)
and solve Eqs. (2) and (3) simultaneously for B and D.
Subtract Eq. (3) from Eq. (2) to give
ln
7.1 10 5
1.9 10
B
5
1
499.67
1
➔ B 3650 K.
609.67
Substitute this value of B into Eq. (2) to yield
ln 7.1 10 5
3650
At T
459.67
80 F
4.77
1
499.67
ln D ➔ D
4.77 10 8
539.67 R
3650
8 539.67
10 e
4.13 10 5
lb s
ft 2
lb s
ft 2
.
Problem 1.72
For a parallel plate arrangement of the type shown in the figure below it is found that when
the distance between plates is 2 mm , a shearing stress of 150 Pa develops at the upper plate
when it is pulled at a velocity of 1 m s. Determine the viscosity of the fluid between the
plates. Express your answer in SI units.
δa
U
P
u
b
B
B'
A
δβ
y
Fixed plate
Solution 1.72
du
U
dy
b
N
m2
m
1
s
0.002 m
150
U
b
0.300
N s
m2
Problem 1.73
Two flat plates are oriented parallel above a fixed lower plate as shown in the figure below.
The top plate, located a distance b above the fixed plate, is pulled along with speed V . The
other thin plate is located a distance cb , where 0 c 1, above the fixed plate. This plate
moves with speed V1, which is determined by the viscous shear forces imposed on it by the
fluids on its top and bottom. The fluid on the top is twice as viscous as that on the bottom.
Plot the ratio V1 V as a function of c for 0 c 1.
V
2μ
b
V1
μ
cb
Solution 1.73
For constant speed, V1, of the middle plate, the net force on the plate is 0, Hence,
Ftop Fbottom where F
A . Thus, the shear stress on the top and bottom of the plate must
be equal.
du
dy
bottom where
top
For the bottom fluid
du
dy
(1)
V1
du
, while for the top fluid
dy
cb
V V1
b1 c
Hence, from Eqn. (1), 2
Note: c
0
V1
V
0, c
V1
cb
V1
V
2c
c 1
2
, c 1
3
V1
V
1
0.6
0.8
2cV1 V1 cV1 ➔
2cV
1
2
V1
V
V V1
.
b cb
1.0
0.8
0.6
V1
V
0.4
0.2
0.0
0.0
0.2
0.4
c
1.0
Problem 1.74
Three large plates are separated by thin layers of ethylene glycol and water, as shown in the
figure below. The top plate moves to the right at 2 m/s. At what speed and in what direction
must the bottom plate be moved to hold the center plate stationary?
Copper plate
Water (20°C)
2 m/s
0.1 cm
Plastic
plate
0.2 cm
Steel plate
Ethylene glycol (20°C)
Solution 1.74
The center plate is stationary if F1
Newtonian fluids and thin layers,
du
dy center
F
F2 (see image). Assuming
V1
μ1
F2
V
h
V2
plate
so
1
V1
h1
2
V2
h2
or
V2
1
2
h2
V1
h1
w
heg
eg
hw
From the liquid properties table:
V2
N s
m2
N s
1.99 10 2 2
m
V2
0.201
1.00 10 3
m
, left.
s
eg
0.2 cm
0.1cm
V1.
1.99 10 2
2
m
s
N s
m
2
and
w
1.00 10 3
N s
m2
.
F1
h1
μ2
h2
Problem 1.75
There are many fluids that exhibit non-Newtonian behavior. For a given fluid, the distinction between Newtonian and non-Newtonian behavior is usually based on measurements of
shear stress and rate of shearing strain. Assume that the viscosity of blood is to be deterdu
mined by measurements of shear stress, , and rate of shearing strain, , obtained from a
dy
small blood sample tested in a suitable viscometer. Based on the data given below, determine if the blood is a Newtonian or non-Newtonian fluid. Explain how you arrived at your
answer.
(N/m 2 )
0.04
0.06
0.12
0.18
0.30
0.52
1.12
2.10
du /dy (s 1 )
2.25
4.50
11.25
22.5
45.0
90.0
225
450
0.0050
0.0047
Solution 1.75
For a Newtonian fluid the ratio of
du /dy
N s/m 2
0.0178
0.0133
to
du
is a constant. For the data given:
dy
0.0107 0.0080
0.0067
0.0058
The ratio is not a constant but decreases as the rate of shearing strain increases. Thus, this
fluid (blood) is a non-Newtonian fluid.
NOTE: The behavior of many non-Newtonian fluids can be well approximated by a power-law relationship. If that is true for this fluid, on a log-log plot the relationship between
shear stress and strain rate should be a straight line.
10.0
1.0
τ (N/m2)
0.1
0.0
1
10
100
1,000
du , 1
dy s
It appears that for this sample, the blood indeed is well represented by a power-law.
Problem 1.76
The sled shown in the figure below slides along on a thin horizontal layer of water between
the ice and the runners. The horizontal force that the water puts on the runners is equal to
1.2 lb when the sled’s speed is 50 ft s. The total area of both runners in contact with the
water is 0.08 ft 2, and the viscosity of the water is 3.5 10 5 lb s ft 2 . Determine the thickness of the water layer under the runners. Assume a linear velocity distribution in the water
layer.
Solution 1.76
F (force)
dv
dy
A
V
where d
d
thickness of water layer.
Thus,
F
V
A
d
and
d
VA
F
3.5 10 5
lb s
ft
50
2
s
ft
1.2 lb
0.08 ft 2
11.7 10 4 ft
Problem 1.77
A 25-mm-diameter shaft is pulled through a cylindrical bearing as shown in the figure
below. The lubricant that fills the 0.3-mm gap between the shaft and bearing is oil
having a kinematic viscosity of 8.0 10 4 m2 s and a specific gravity of 0.91. Determine the
force P required to pull the shaft at a velocity of 3 m s. Assume the velocity distribution in
the gap is linear.
Bearing
Lubricant
P
Shaft
0.5 m
Solution 1.77
τA
P
D
Fx
x
0
Thus,
P
A
where A
D
shaft length in bearing
and
velocity of shaft
gap width
D
V
b
so that
V
b
P
v
v SG
8.0 10 4
m2
s
Since
P
P
V
b
D
286
k m 1 N s2
s2 1 kg m
D
H2O @ 4 C ,
kg
m
3
3
s
m
0.0003 m
0.91 103
286 N
0.025 m 0.5 m
Problem 1.78
A hydraulic lift in a service station has a 32.50-cm- diameter ram that slides in a 32.52-cmdiameter cylinder. The annular space is filled with SAE 10 oil at 20°C . The ram is traveling
upward at the rate of 0.10 m/s. Find the frictional force when 3.0 m of the ram is engaged in
the cylinder.
Solution 1.78
0.10 m/s
Modeling the oil as a Newtonian fluid:
du
dy
H=3m
linear velocity profile aross gap.
2R
du
dy
0.10
0.01cm
m
s
m
100 cm
1
1000 .
s
0.01 cm
gap
oil
32.52 cm
0.123Pa s at 20 o C
0.123
N s
m2
1000
F
A
F
377 N
Therfore,
1
s
2 RH
123
N
m2
123
N
m
2
2
16.25 cm 3 m
1m
100 cm
Problem 1.79
A piston having a diameter of 5.48 in. and a length of 9.50 in. slides downward with a velocity V through a vertical pipe. The downward motion is resisted by an oil film between the
piston and the pipe wall. The film thickness is 0.002 in., and the cylinder weighs 0.5 lb. Estimate V if the oil viscosity is 0.016 lb s ft 2 . Assume the velocity distribution in the gap is
linear.
Solution 1.79
τ
𝒲
ℓ
τA
D
Constant velocity
Linear velocity profile
Fvertical
0
W
A
D
Newtonian fluid
du
dy
0.002
ft
12
9.50
lb s
ft 0.016 2
12
ft
0.00459
velocity
film thickness
Substitution yields:
V
W
V
W
D
D
0.5 lb
5.48
ft
12
ft
s
V
Problem 1.80
A 10-kg block slides down a smooth inclined surface as shown in the figure below. Determine the terminal velocity of the block if the 0.1-mm gap between the block and the surface
contains SAE 30 oil at 60 F. Assume the velocity distribution in the gap is linear, and the
area of the block in contact with the oil is 0.1 m2 .
0.1 mm gap
V
20°
Solution 1.80
Draw free body diagram to help resolve forces:
y
V
x
Constant velocity
W
Fx
0
W sin
A
τA
N
Linear velocity profile
velocity
film thickness
du
dy
Newtonian fluid
where b is film thickness. Substitution yields:
V
W sin
mg sin
A
b
V
V
bmg sin
A
0.0883
m
s
0.0001 m 10 kg 9.81
0.38
N s
m2
m
s2
0.1 m 2
sin 20
1 N s2
1 kg m
V
b
Problem 1.81
A layer of water flows down an inclined fixed surface with the velocity profile shown in the
figure below. Determine the magnitude and direction of the shearing stress that the water
exerts on the fixed surface for U 2 m s and h 0.1 m .
u
U
h
y
2
h
y2
h2
U
u
y
Solution 1.81
Enforcing the no-slip boundary condition at the solid surface:
du
dy
y2
d
y
U 2
dy
h
Thus, at the fixed surface y
u
2D
y y 0
h
U
h2
2
h
2y
h2
0
Thus
y 0
U
2
h
1.12 10 3
4.48 10 2
N
m2
N s
m
2
2
m
2
s 0.1 m
acting in direction of flow
Problem 1.82
Oil (absolute viscosity 0.0003lb s/ft 2, density 50 lbm/ft3) flows in the boundary layer, as
shown in the figure below. The plate is 1 ft wide perpendicular to the paper. Calculate the
shear stress at the plate surface.
Edge of boundary layer
y
u
u = 1500 y – 5 × 106 × y3 ft/sec
0.01 ft
Plate
Solution 1.82
Assuming a newtonian fluid, the shear stress on the plate by the oil is
du
dy y 0
For the given velocity profile:
u
5 10 6
1500
y
sec
sec ft
du
dy
1500 15 106 y2
➔
sec
sec ft 2
1
sec
0.45
y3 ➔
2
Substitution yields:
0.0003lb sec
ft
2
1500
lb
s
du
1
1500
dy y 0
sec
Problem 1.83
Standard air flows past a flat surface, and velocity measurements near the surface indicate
the following distribution:
y ft
0.005
0.01
0.02
0.04
0.06
0.08
u ft/s
0.74
1.51
3.03
6.37
10.21
14.43
The coordinate y is measured normal to the surface and u is the velocity parallel to the surface. (a) Assume the velocity distribution is of the form u C1 y C2 y3 and use a standard
curve-fitting technique to determine the constants C1 and C 2.
(b) Make use of the results of part (a) to determine the magnitude of the shearing stress at
the wall (y 0) and at y 0.05 ft .
Solution 1.83
(a) Use nonlinear regression program to obtain coefficients C1 and C 2. The program produces least squares estimates of the parameters of a nonlinear model. For the data
given,
C1 153 s-1 and C2 4350 ft -2s-1
(b) For a Newtonian fluid:
du
d
C1 y 3C2 y2
dy
dy
At the locations specified:
y 0
y 0.05 ft
C1
3.74 10 7
3.74 10 7
lb s
ft
2
C1 6 C2 y
lb s
ft
2
153
153
1
s
5.72 10 5
1
1
3 4350 2
s
ft s
lb
ft 2
0.05 ft
2
6.94 10 5
lb
ft 2
Problem 1.84
A new computer drive is proposed to have a disc, as shown in the figure below. The disc is
to rotate at 10,000 rpm , and the reader head is to be positioned 0.0005 in. above the surface
of the disc. Estimate the shearing force on the reader head as a result of the air between the
disc and the head.
Stationary reader head
0.2-in.dia.
10,000 rpm
0.0005 in.
2 in.
Rotating disc
Solution 1.84
F
shear force on head
A
Assuming a uniform and linear velocity profile:
du
dy
U
b
3.74 10 7
lb s
ft 2
0.0005
ft
12
F
A
1.57
R
b
lb
ft
2
10000
0.2
ft
4 12
rev 1 min 2 rad
min 60 s
1 rev
2
3.43 10 4 lb
2
ft
12
1.57
lb
ft 2
Problem 1.85
The space between two 6-in.-long concentric cylinders is filled with glycerin
(viscosity 8.5 10 3 lb s ft 2 ). The inner cylinder has a radius of 3 in. and the gap width
between cylinders is 0.1 in. Determine the torque and the power required to rotate the inner
cylinder at 180 rev min. The outer cylinder is fixed. Assume the velocity distribution in the
gap to be linear.
Solution 1.85
The torque on a sector of the cylinder surface corresponding
to an included angle of dq is:
dT
dFRi
Liquid
dARi
Ri d
𝒯
ω
2
Ri
Fixed
outer
cylinder
Ri d
Integrating around the cylinder:
2
Ri2
T
d
Rotating
inner
cylinder
Ri2
2
0
For a linear velocity distribution in the gap
Ri
R0 Ri
T
Ri
Ri2
R0 Ri
2
Ri
Ro
Ri3
2
R0 Ri
τ
180
rev
min
2
rad
rev
1min
60 s
6
rad
s
Ri
dθ
Ro
2
T
3
ft
12
3
6
ft
12
8.5 10 3
lb s
ft 2
6
rad
s
(ℓ ~ cylinder length)
0.1
ft
12
T
0.944 ft lb
To compute power requied:
power T
= 0.944 ft lb 6
top view
rad
s
17.8
ft lb
s
ℓ
Problem 1.86
A pivot bearing used on the shaft of an electrical
instrument is shown in the figure below. An oil with
5000 rpm
2
0.010 lb s ft fills the 0.001-in.
a viscosity of
gap between the rotating shaft and the stationary
base. Determine the frictional torque on the shaft
when it rotates at 5000 rpm .
0.2 in.
30°
μ = 0.010 lb • s/ft2
0.001 in.
Solution 1.86
torque on area element dA, where dA 2 rd
Let dT
2 rdr
sin
Thus,
d
rdF
du
dy
r dA where
ω
r
b
R
so that,
r
d
r
b
r
2 rdr
sin
dℓ
2
r3dr
b sin
dA
b
Hence,
2
b sin
d
r R
r3dr
r 0
2b sin
R4
(1)
dr
θ
Now,
R
0.1 in., b 0.001 in.,
lb s
ft
rad
rev
524
lb s
rad
524
2
s
ft
0.001
2
ft sin30
12
0.1
ft
12
5000
rev min
min 60 s
0.010
2
2
,
30 deg, and
rad
s
Thus, from Eq. (1),
0.010
4
9.53 10 4 ft lb
dr
dℓ = sinθ
Problem 1.87
The viscosity of liquids can be measured through the use of a
rotating cylinder viscometer of the type illustrated in the figure
below. In this device the outer cylinder is fixed and the inner
cylinder is rotated with an angular velocity, . The torque
required to develop is measured and the viscosity is calculated
from these two measurements. (a) Develop an equation relating
, , , , R0, and Ri . Neglect end effects and assume the velocity
distribution in the gap is linear. (b) The following torque-angular
velocity data were obtained with a rotating cylinder viscometer of
the type discussed in part (a).
𝒯
Liquid
Fixed
outer
cylinder
ω
Ri
Ro
Rotating
inner
cylinder
Torque (ft lb)
13.1
26.0
39.5
52.7
64.9
78.6
Angular velocity
(rad/s)
1.0
2.0
3.0
4.0
5.0
6.0
ℓ
For this viscometer R0 2.50 in. , Ri 2.45 in. , and
5.00 in. Make use of these data and
a standard curve-fitting program to determine the viscosity of the liquid contained in the
viscometer.
Solution 1.87
(a) Torque, d , on infinitesimal-size axial strip on the
inner cylinder surface due to shearing stress:
d
Ri dA Ri Ri d
Ri2 d
2
Integration yields:
Ri2
d
2 Ri2
0
For a linear velocity distribution in the gap:
Ri
R0 Ri
2 Ri3
R0 Ri
(1)
τ
Ri
dθ
Ro
top view
(ℓ ~ cylinder length)
(b) For a fixed geometry and a given viscosity, Eq. (1) is of the form
2 Ri3
y bx where: y
, x
, b
R0 Ri
Entering the data provided into Excel: b 13.08 ft lb s
Solving for viscosity:
b R0 Ri
2
Ri3
13.08 ft lb s
2
2.45
ft
12
2.50 2.45
ft
12
3
5.00
ft
12
2.45
lb s
ft 2
Problem 1.88
3.0 cm
The concentric cylinder viscometer shown in the figure
below has a cylinder height of 10.0 cm , a cylinder radius
of 3.0 cm , and a uniform gap between the cylinder and
the container (bottom and sides) of 0.10 cm . The pulley
has a radius of 3.0 cm. Determine the weight required to
produce a constant rotational speed of 30 rpm if the gap
is filled with: (a) water, (b) gasoline, (c) glycerin.
𝒲
H = 10.0 cm
0.10 cm
0.10 cm
Solution 1.88
u = rω
Resisting torque is due to shear stress acting on cylinder surfaces.
Assuming a linear velocity profile across the narrow gaps, the torque
on the cylinder wall is:
du
dy w
A R
1
R
h
DH R
2 RH R
ω
R
HR3
h
2
y
The velocity at the cylinder bottom is a function of radial position.
The infinitesimal torque acting on an annular ring of differential
width is:
d 2
T2
r
h
dA r
2
h
2 rdr r
R 3
r dr
0
2h
1
2
Total torque:
2
W
( a ) water
rev
min
2
1.910
R3
H
h
W
R 4 2H
hRp
R
1.910
(c ) glycerine
W
1.910
velocity
profile,
u = rω
4
2
2
2 10 cm
m2
s
N s
m
3.10 10 4
1.50
N s
m2
1
2
1m
100 cm
2.14 10 3 N
2
N s
m
1
2
1
2
3 cm
1.12 10 3
m
s
W
2h
3 cm
R 4 2H
h
R
R4
3 cm
m2
s
( b ) gasoline
r3dr
h
2 rad 1 min
1 rev 60 sec
0.1 cm
W
zero
velocity
R4
Neglecting friction: WRp
30
r
h
2
5.92 10 4 N
2.87 N
2
1.91
m2
s
Problem 1.89
Rotating plate
A 12-in.-diameter circular plate is placed
Torque
over a fixed bottom plate with a 0.1-in.
0.1-in. gap
gap between the two plates filled with
glycerin as shown in the figure below.
Determine the torque required to rotate
the circular plate slowly at 2 rpm . Assume that the velocity
distribution in the gap is linear and that the shear stress on the edge of the rotating plate
is negligible.
Solution 1.89
As shown, considering an annular ring of differential width
d
r dA
τ
r 2 rdr
R
r 2 dr
2
Integration yields:
dr
r
0
For the annular strip:
Thus,
2
R
r
R
r3dr
0
2
Stresses acting on bottom of plate
R4
4
V = rω
δ
Using the data specified:
2
T
lb s
0.0313 2
ft
0.0772 ft lb
rev
rad
2
2
min
rev
0.1
ft 4
12
1min
60 s
6
ft
12
y
u
4
du
V =—
rω–
__ = __
δ
dy δ
Velocity distribution
Problem 1.91
Some measurements on a blood sample at 37 °C (98.6 °F) indicate a shearing stress of
0.52 N m2 for a corresponding rate of shearing strain of 200 s 1. Determine the apparent
viscosity of the blood and compare it with the viscosity of water at the same temperature.
Solution 1.91
du
dy
N
m 2 26.0 10 4 N s
blood
1
m2
200
s
From Table B.2 Physical Properties of Water (SI Units)
N s
@30 C H2O 7.975 10 4 2
m
4 N s
@ 40 C H2O 6.529 10
m2
0.52
Thus, with linear interpolation,
and
N s
m2
N s
6.96 10 4 2
m
26.0 10 4
blood
H 2O
3.74
H2O
37 C
6.96 10 4
N s
m2
Problem 1.93
A sound wave is observed to travel through a liquid with a speed of 1500 m s. The specific
gravity of the liquid is 1.5. Determine the bulk modulus for this fluid.
Solution 1.93
c
EN
Thus,
EN c2
SG H2O and SG
, where
c 2SG H 2O
m
1500
s
3.37 109
2
1.5 999
kg m
s2 m 2
or
EN
3.37 109
N
m2
kg
m3
1.5
Problem 1.94
A rigid-walled cubical container is completely filled with water at 40 °F and sealed. The water is then heated to 100 °F. Determine the pressure that develops in the container when the
water reaches this higher temperature. Assume that the volume of the container remains
constant and the value of the bulk modulus of the water remains constant and equal to
300,000 psi.
Solution 1.94
Since the water mass remains constant,
40
100
where
is volume and
Thus,
40
is change in volume if water were unconstrained during heating.
1
100
From the table Physical Properties of Water (BG/EE Units)
40
1.940
slugs
ft
3
and
100
1.927
slugs
ft3
so that
slugs
ft3 1 0.00675
slugs
1.927 3
ft
dp
it follows with d
From the equation E
1.940
and dp
sure required to compress the water back to its original volume is
p
300000 psi 0.00675
2.03 103 psi
p that the change in pres-
Problem 1.95
Estimate the increase in pressure (in psi) required to decrease a unit volume of mercury
by 0.1 %.
Solution 1.95
dp
Ev
d
p
Thus,
p
p
Ev
4.14 103 psi
4.14 106
lb
in.2
0.001
Problem 1.96
A 1m3 volume of water is contained in a rigid container. Estimate the change in the volume
of the water when a piston applies a pressure of 35 MPa .
Solution 1.96
dp
Ev
d
Thus,
p
N
p
m2
N
Ev
2.15 109 2
m
or decrease in volume 0.0163 m3
1 m3
35 106
0.0163 m3
Problem 1.97
Determine the speed of sound at 20 °C in (a) air, (b) helium, and (c) natural gas (methane).
Express your answer in m s.
Solution 1.97
c
kRT
With T
20°C 273 293K:
(a) For air, c
1.40 286.9
(b) For helium, c
(c) For natural gas, c
J
kg K
1.66 2077
293K
J
kg K
1.31 518.3
1N m
1J
293K
J
kg K
1 kg m
1N s
1N m
1J
293K
343
2
1 kg m
1N s
1N m
1J
2
m
s
1010
1 kg m
1 N s2
m
s
446
m
s
Problem 1.98
Air is enclosed by a rigid cylinder containing a piston. A pressure gage attached to the cylinder indicates an initial reading of 25 psi. Determine the reading on the gage when the piston has compressed the air to one-third its original volume. Assume the compression process to be isothermal and the local atmospheric pressure to be 14.7 psi.
Solution 1.98
For isothermal compression,
p
constant=
pi
pf
i
f
where i  initial state and
f  final state.
Thus, p f
f
pi
i
Because the mass of air is constant:
Therefore,
pf
3 25 14.7 psi (abs)
mass
volume
119 psi (abs)
or
p f (gage)
119 14.7 psi 104 psi (gage)
f
i
initial volume
final volume
3.
Problem 1.99
Air is enclosed by a rigid cylinder containing a piston. A pressure gage attached to the cylinder indicates an initial reading of 25 psi. Determine the reading on the gage when the piston has compressed the air to one-third its original volume. Assume the compression process takes place without friction and without heat transfer (isentropic process) and the local
atmospheric pressure to be 14.7 psi.
Solution 1.99
For isentropic compression,
p
k
constant=
pi
pf
k
i
k
f
where i  initial state and
f  final state.
Thus p f
f
pi
i
Because the amount of mass is constant:
mass
volume
f
i
initial volume
final volume
Therefore,
pf
3
1.40
25 14.7 psi (abs)
184.8psi (abs)
or
p f (gage) 184.8 14.7 170 psi (gage)
3
Problem 1.100
Carbon dioxide at 30 °C and 300 kPa absolute pressure expands isothermally to an absolute pressure of 165 kPa . Determine the final density of the gas.
Solution 1.100
For isothermal compression,
p
constant=
pi
pf
i
f
where i  initial state and
f  final state.
pf
Thus, f
i
pi
Also,
pi
RTi
i
300 103
188.9
J
1N m
kg K
1J
N
m2
30 °C 273 K
Therefore,
f
165 kPa
300 kPa
5.24
kg
m
3
2.88
kg
m3
5.24
kg
m3
Problem 1.101
Natural gas at 70 °F and standard atmospheric pressure of 14.7 psi (abs) is compressed
isentropically to a new absolute pressure of 70 psi. Determine the final density and
temperature of the gas.
Solution 1.101
p
For isentropic compression,
k
constant=
pi
pf
k
i
k
f
where i  initial state and
f  final state.
Therefore
Also,
i
Therefore,
pf
f
pi
pi
RTi
f
1
k
i
lb
14.7 2
in.
3.099 103
ft lb
slug °R
1
1.31
70 psia
14.7 psia
in.2
144 2
ft
1.29 10 3
70 °F 460 °R
1.29 103
slugs
ft
3
4.25 10 3
slugs
ft3
slugs
ft3
Using the ideal gas model:
Tf
70
pf
fR
4.25
lb
in.2
slugs
10 3 3
ft
144
in.2
ft3
ft lb
3.099 10
slug °R
3
765°R
765 460 °F 305°F
Problem 1.102
A compressed air tank in a service station has a volume of 10 ft3. It contains air at 70°F and
150 psia. How many tubeless tires can it fill to 44.7 psia at 70 °F if each tire has a volume of
1.5 ft 3 and the compressed air tank is not refilled? The tank air temperature remains
constant at 70 °F because of heat transfer through the tank’s large surface area.
Solution 1.102
Modelling the air as an ideal gas, the mass of air mt that can be put into each tire is found
from:
mtire
p
p
f
p
RT
i
1.5 ft3 44.7 14.7
mtire
ft lb
1716
slug o R
p
RT i
f
lb
in.
12
ft
in.2
RT
pf
pi
2
0.007125slug.
530 o R
Air in the tank can be put into the tires until the tank air pressure drops to 44.7
lb
in.2
lute. The mass of air mT that can be taken out of the tank and put into the tires is
mT
p
i
p
f
RT
10 ft3 150 44.7
ft lb
1716
slug o R
pi
pf
lb
in.
12
ft
in.2
2
0.1667 slug.
530 o R
The number of tires that can be filled is
No.
0.1667 slug
0.007125slug
23.4 or
No. 23tires
abso-
Problem 1.103
A regulation basketball is initially flat and is then inflated to a pressure of approximately
24 lb/in 2 absolute. Consider the air temperature to be constant at 70 °F . Find the mass of
air required to inflate the basketball. The basketball’s inside radius is 4.67 in.
Solution 1.103
Modelling air as an ideal gas and looking up the gas constant for air:
p
RT
p 4
Ri3
RT 3
lb
in.2
4
3
1716
ft lb
slug o R
m
24
m 0.000938slug.
4.67 in.
3
ft
12 in.
460 70 o R
Problem 1.104
Assume that the air volume in a small automobile tire is constant and equal to the volume
between two concentric cylinders 13 cm high with diameters of 33 cm and 52 cm. The air in
the tire is initially at 25 °C and 202 kPa. Immediately after air is pumped into the tire, the
temperature is 30 °C and the pressure is 303 kPa . What mass of air was added to the tire?
What would be the air pressure after the air has cooled to a temperature of 0 °C ?
Solution 1.104
The mass of air added to the tire is the difference of the final mass of
air m f and the initial mass mi . Assuming air is an ideal gas,
mf
p
RT
mi
f
p
RT i
pf
pi
.
Ti
R Tf
r1
r2
h
h = 13 cm
d1 = z r1 = 33 cm
d2 = z r2 = 52 cm
Using specified data:
r22
r12
h
26 cm
2
0.0165 m3
mf
mi
287.0
mf
mi
N m
kg K
16.5 cm
2
13 cm
303kPa
273 30 K
m
100 cm
202 kPa
273 25 K
3
0.0165 m 3 .
N
m2
kPa
1000
0.0185kg
Now consider the cooling process. The initial state will be 30°C and 303kPa . The final state
will be 0°C and p f . Applying the ideal gas law to both states gives
p
RT i
Since
f
i,
pf
pi
p
RT
Tf
Ti
.
f
303 kPa
273 0
273 30
pf
273 kPa
Problem 1.105
Develop a computer program for calculating the final gage pressure of gas when the initial
gage pressure, initial and final volumes, atmospheric pressure, and the type of process (isothermal or isentropic) are specified. Use BG units. Check your program against the results
obtained for Problem 1.98.
Solution 1.105
For compression or expansion,
p
n
constant ,
where n 1 for an isothermal process, and n specific heat ratio for an isentropic process.
pf
p
Therefore: ni
where i  initial state and f  final state, so that
n
i
pf
f
n
f
pi
(1)
i
Because the amount of mass is constant: m
i
i
f
f
i
i
f
f
Recognizing that the pressure in Eq. 1 must be absolute pressure:
n
p fg
i
patm
pig
patm
(2)
f
where the subscript g refers to gage pressure. Equation (2) can be written as
p fg
Vi
Vf
n
pig
patm
patm
(3)
A spreadsheet (EXCEL) program for calculating the final gage pressure follows.
This program calculates the final gage pressure of an ideal gas when the
initial gage pressure in psi, the initial volume, the final volume, the
atmospheric pressure in psia, and the type of process (isothermal or
isentropic) is specified. To use, replace current values and let k=1 for
isothermal process or k=specific heat for isentropic process.
A
B
C
D
E
F
Final
Initial gage
Initial
Final
Atmospheric
gage
pressure
volume
volume
pressure
pressure
Pig(psi)
25
Vi
1
Vf
0.3333
Patm(psia)
14.7
k
1
Pfg(psia)
104.4
Row 10
Formula:
=((B10/C10)^E10)*(A10+D10)-D10
Data from Problem 1.98 are included in the above table, giving a final gage
pressure of 104.4 psi.
Problem 1.106
Often the assumption is made that the flow of a certain fluid can be considered as incompressible flow if the density of the fluid changes by less than 2 %. If air is flowing through a
tube such that the air pressure at one section is 9.0 psi and at a downstream section it is 8.6
psi at the same temperature, do you think that this flow could be considered an incompressible flow? Support your answer with the necessary calculations. Assume standard atmospheric pressure.
Solution 1.106
Modelling the air as an ideal gas undergoing an isothermal process:
p1 p2
p2
2
p
RT
p1
1
2
1
% density change
1
2
100
2
100
1
1
1
1
p2
p1
100
1
8.6 14.7 psia
9.0 14.7 psia
100
1.69 % 2 %
Yes. This process is well modelled as an incompressible flow.
Problem 1.107
An important dimensionless parameter concerned with very high-speed flow is the Mach
number, defined as V c, where V is the speed of the object such as an airplane or projectile,
and c is the speed of sound in the fluid surrounding the object. For a projectile traveling at
800 mph through air at 50 °F and standard atmospheric pressure, what is the value of the
Mach number?
Solution 1.107
V
c
From the table of Physical Properties of Air at Standard Atmospheric Pressure (BG/EE
Units)
Mach number
cair @50°F
1106
ft
s
ft
mi
ft
1106
s
800 mph 5280
Mach number=
1 hr
3600 s
1.06
Problem 1.108
The “power available in the wind” of velocity V through an area A is
W
1
AV 3,
2
where is the air density (0.075 lbm/ft3). For an 18-mph wind, find the wind area A that will
supply a power of 4 hp .
Solution 1.108
Solving for the area A and using appropriate unit conversion factors:
A
2 4 hp
2W
V3
lb
0.075 m
ft3
3.17
A 102 ft 2
mi
18
hr
550 ft lb/s
1 hp
3
5280 ft 1 hr
1 mi
3600 s
ft lb s2 1 lbm 32.2 ft
lb m
1 lb s2
3
,
Problem 1.109
Air enters the converging nozzle shown in the figure below at T1
At the exit of the nozzle, V2 is given by
V2
where c p
V12
V1
2c p T1 T2 ,
1000 ft/s .
V2
2
1
Solution 1.109
Solving for T2 and inserting the values specified: qc gives
T2
50 ft/s .
187 ft lb/lbm °F and T2 is the air temperature at the exit of the nozzle. Find the
temperature T2 for which V2
T2
70 °F and V1
T1
V22 V12
2c p
12.8 o F
1000
70 o F2 187
ft
s
ft lb
lbm o F
2
50
ft
s
2
1 lb m 32.2 ft
1 lb s2
Problem 1.110
This water jet is a blast Usually, liquids can be treated as incompressible fluids. However, in
some applications the compressibility of a liquid can play a key role in the operation of a
device. For example, a water pulse generator using compressed water has been developed
for use in mining operations. It can fracture rock by producing an effect comparable to a
conventional explosive such as gunpowder. The device uses the energy stored in a waterfilled accumulator to generate an ultrahigh-pressure water pulse ejected through a 10-to
25-mm-diameter discharge value. At the ultrahigh pressures used (300 to 400 MPa, or
3000 to 4000 atmospheres), the water is compressed (i.e., the volume reduced) by about
10% to 15%. When a fast-opening valve within the pressure vessel is opened, the water expands and produces a jet of water that upon impact with the target material produces an
effect similar to the explosive force from conventional explosives. Mining with the water jet
can eliminate various hazards that arise with the use of conventional chemical explosives,
such as those associated with the storage and use of explosives and the generation of toxic
gas by-products that require extensive ventilation. (See Problem 1.110.)
By what percent is the volume of water decreased if its pressure is increased to an equivalent
to 3000 atmospheres (44,100 psi)?
Solution 1.110
EV
dp
d
p
p
44100 psia 14.7 psia
EV
3.12 105 psia
Thus, % decrease in volume = 14 %
0.141
Problem 1.111
During a mountain climbing trip it is observed that the water used to cook a meal boils at
90 °C rather than the standard 100 °C at sea level. At what altitude are the climbers preparing their meal? See Table of Physical Properties of Water (SI Units) and Table of Properties
of the U.S. Standard Atmosphere (SI Units) for data needed to solve this problem.
Solution 1.111
Water boils when the vapor pressure of the liquid is the same as atmospheric pressure.
N
From the water property table, at T 90°C, pv 7.01 104 2 (abs).
m
N
From standard atmosphere table, p 7.01 10 4 2 (abs)
altitude 3000 m
m
Problem 1.112
When a fluid flows through a sharp bend, low pressures may develop in localized regions of
the bend. Estimate the minimum absolute pressure (in psi) that can develop without causing
cavitation if the fluid is water at 160 °F.
Solution 1.112
Cavitation may occur when the local pressure equals the vapor pressure. For water at 160 °F
(from Table of Physical Properties of Water [BG/EE Units])
T
160 o F
pv
4.74 psi (abs)
minimum pressure
4.74 psia
Problem 1.113
A partially filled closed tank contains ethyl alcohol at 68 °F. If the air above the alcohol is
evacuated, what is the minimum absolute pressure that develops in the evacuated space?
Solution 1.113
Minimum pressure = vapor pressure = 0.85 psi (abs)
Problem 1.114
Estimate the minimum absolute pressure (in Pascals) that can be developed at the inlet of a
pump to avoid cavitation if the fluid is carbon tetrachloride at 20 °C .
Solution 1.114
Cavitation may occur when the section pressure at the pump inlet equals the vapor pressure
For carbon tetrachloride at 20 °C , pv
13 kPa (abs) .
Thus, minimum pressure = 13 kPa (abs)
Problem 1.115
When water at 70 °C flows through a converging section of pipe, the pressure decreases in
the direction of flow. Estimate the minimum absolute pressure that can develop without
causing cavitation. Express your answer in both BG and SI units.
Solution 1.115
Cavitation may occur in the converging section of pipe when the pressure equals the vapor
pressure. From the Table of Physical Properties of Water (SI Units) for water at 70 °C ,
pv 31.2 kPa (abs).
Therefore,
minimum pressure 31.2 kPa (abs)
3
31.2 10
4.52 psia
N
m2
1.450 10 4 psi
N
1 2
m
Problem 1.116
At what atmospheric pressure will water boil at 35 °C ? Express your answer in both SI and
BG units.
Solution 1.116
The vapor pressure of water at 35°C is 5.81 kPa(abs) (from Table of Physical Properties of
Water [SI Units] using linear interpolation).
Thus, if water boils at this temperature the atmospheric pressure must be equal to
5.81kPa (abs) in SI units. In BG units,
5.81 103
N
m2
1.450 10 4 psi
N
1 2
m
0.842 psi (abs)
Problem 1.118
When a 2-mm-diameter tube is inserted into a liquid in an open tank, the liquid is observed
to rise 10 mm above the free surface of the liquid. The contact angle between the liquid and
the tube is zero, and the specific weight of the liquid is 1.2 104 N m3 . Determine the value
of the surface tension for this liquid.
Solution 1.118
θ
2π Rσ
θ
h
γπ R2h
h
2R
The action of surface tension in a tube inserted into a pool is to draw upward (or depress)
2 cos
the liquid in the tube a distance h
with respect to the elevation of the surrounding
R
free surface.
For the specified contact angle, = 0:
N
1.2 10
10 10 3 m
3
m
2 cos 0
4
hR
2 cos
2 10 3 m
2
0.060
N
m
Problem 1.119
A soda straw with an inside diameter of 0.125 in. is inserted into a pan of water at 60 °F . The
water in the straw rises to a height of 0.150 in. above the water surface in the pan. Determine
the value of , the contact angle of the water with the straw (see the figure below).
θ
2π Rσ
θ
γπ R2h
h
h
2R
(a)
(b)
(c)
Effect of capillary section in small tubes. (a) Rise of column for a liquid that wets the tube, (b) Free-body diagram
for calculating column height. (c) Depression of column for a nonwetting liquid.
Solution 1.119
Consider the free body diagram of the liquid column inside the tube as shown in the figure.
If the liquid column is in static equilibrium:
F
0
2 R cos
R2 h
cos
R2 h
2 R
cos 1
Rh
2
lb
62.4 3
ft
0.0625
ft
12
2 5.04 10 3
1.16 radians 66.2o
0.150
ft
12
lb
ft
Problem 1.120
Small droplets of carbon tetrachloride at 68 °F are formed with a spray nozzle. If the average diameter of the droplets is 200 m, what is the difference in pressure between the inside
and outside of the droplets?
Solution 1.120
From the force balance on a half-droplet presented in the chapter:
2
p
R
Looking in the properties table for carbon tetrachloride at 68 °F,
Substitution yields:
N
m
6
100 10 m
2 2.69 10 2
p
538Pa
2.69 10 2
N
.
m
Problem 1.121
A 12-mm-diameter jet of water discharges vertically into the atmosphere. Due to surface
tension, the pressure inside the jet will be slightly higher than the surrounding atmospheric
pressure. Determine this difference in pressure.
Solution 1.121
Considering the free body diagram of one half of a short
length of jet, , equilibrium requires:
R
p 2R
2
p
δℓ
R
7.34 10 2
N
m
1 12
m
2 1000
12.2 Pa
σ
p2Rδ ℓ
p~excess pressure
Surface tension force = σ 2 δ ℓ
Problem 1.122
A method used to determine the surface tension of a liquid is to determine the force necessary to raise a wire ring through the air–liquid interface, as shown in the figure below. What
is the value of the surface tension if a force of 0.015 N is required to raise a 4-cm-diameter
ring? Consider the ring weightless, as a tensiometer (used to measure the surface tension)
“zeroes” out the ring weight.
Solution 1.122
A free body diagram of the ring and supporting wires
is shown on the right and gives
F
F = 0.015 N
2 D
F
2 D
0.015 N
4
2
m
100
5.97 10 2
N
m
4 cm
Water
F
2π Dσ
Problem 1.123
Calculate the pressure difference between the inside and outside of a spherical water droplet
1
having a diameter of in. and a temperature of 50°F.
32
Solution 1.123
A force balance on the outside surface of the drop gives
pi
patm
F
patm R 2
0,
pi R2 2 R
0
pi
2
R
For water at 50°F,
patm
5.09 10 3
2 5.09 10 3
pi
patm
2π Rσ
R
lb
so
ft
lb 1 ft
ft 12 in.
1
in.
32
0.0271 psi
Problem 1.124
Surface tension forces can be strong enough to allow a double-edge steel razor blade to
“float” on water, but a single-edge blade will sink. Assume that the surface tension forces
act at an angle relative to the water surface as shown in the figure below. (a) The mass of
the double-edge blade is 0.64 10 3 kg , and the total length of its sides is 206 mm . Determine the value of required to maintain equilibrium between the blade weight and the resultant surface tension force. (b) The mass of the single-edge blade is 2.6 10 3 kg , and the
total length of its sides is 154 mm . Explain why this blade sinks. Support your answer with
the necessary calculations.
Surface tension
force
θ
Blade
Solution 1.124
(a) water ➔
7.34 10 2
Fvertical
T sin
sin
W
T
N
m
W
T
T
θ
𝒲
0
mg
L
sin 1
m
0.64 kg 9.81 2
s
N
7.34 10 2
0.206 m
m
0.415 radians
24.5o
(b) For the single-edge blade
W
mblade g
m
2.61 10 3 kg 9.81 2
s
0.0256 N
and
T sin
L sin
7.34 10 2
0.0113 N sin
For static equilibrium:
0.0256 N
sin
1, but sin
0.0113 N
N
m
1
0.154 m sin
single-edge blade cannot float
Problem 1.125
Explain how sweat soldering of copper pipe works from a fluid mechanics viewpoint.
Solution 1.125
Solder for sweat soldering copper pipe is an alloy with a melting point below that of copper.
The copper parts are typically heated using a gas torch to a temperature below the melting
point of copper but above the melting point of the solder. When the solder is “touched” to
the joint, it melts. To form a good quality joint between a copper pipe and fittings, or between fittings, capillary action must draw liquid solder into the small gap to between the
two parts to fill the gap and the solder must bond with the copper surface.
From “a fluid mechanics viewpoint,” the flux used for sweat soldering of copper pipe reduces the surface tension of the liquefied solder, reducing the contact angle at the soldercopper interface, thereby producing a stronger capillary action that more effectively draws
the liquid solder into the joint to fill it with solder.
From “a chemical and mechanical viewpoint,” at the elevated temperatures occurring during the soldering process, oxides quickly form on the surface of copper and interfere with
the bonding process. Therefore, even after mechanical cleaning of the parts, flux acts as a
reducing agent to remove oxides from the surface of the copper, facilitating a stronger bond
between the solder and the copper.
Problem 1.126
Under the right conditions, it is possible, due to surface tension, to have metal objects float
490 lb ft3 ) rod on a
on water. Consider placing a short length of a small diameter steel (
surface of water. What is the maximum diameter that the rod can have before it will sink?
Assume that the surface tension forces act vertically upward. Note: A standard paper clip
has a diameter of 0.036 in. Partially, unfold a paper clip and see if you can get it to float on
water. Do the results of this experiment support your analysis?
Solution 1.126
σℓ
In order for rod to float, (see free body diagram):
2
W
4
D2
steel
𝒲
Thus, for the limiting case
2
2
Dmax
4
σℓ
8
ℓ = rod length
steel
steel
so that
8 5.03 10
Dmax
3 lb
lb
490 3
ft
ft
1
2
5.11 10 3 ft
0.0614 in.
Since a standard steel paper clip has a diameter of 0.036 in., which is less than 0.0614 in., it
should float. A simple experiment will verify this. Yes.
Problem 1.127
An open, clean glass tube, having a diameter of 3 mm , is inserted vertically into a dish of
mercury at 20 °C . How far will the column of mercury in the tube be depressed?
Solution 1.127
θ
2π Rσ
θ
γπ R2h
h
h
2R
The action of surface tension in a tube inserted into a pool is to draw upward (or depress)
2 cos
the liquid in the tube a distance h
with respect to the elevation of the surrounding
R
free surface.
For the specified information:
2 4.66 10 1
h
N
133 103 3
m
N
cos130
m
3.00 10 3 m
0.0015 m
Thus, column will be depressed 3.00 mm
Problem 1.128
An open, clean glass tube ( 0°C) is inserted vertically into a pan of water. What tube diameter is needed if the water level in the tube is to rise one tube diameter (due to surface
tension)?
Solution 1.128
θ
2π Rσ
θ
γπ R2h
h
h
2R
The action of surface tension in a tube inserted into a pool is to draw upward (or depress)
2 cos
with respect to the elevation of the surrounding
the liquid in the tube a distance h
R
free surface.
For the specified information:
2R
R2
2 cos 0o
R
5.03 10 3
lb
62.4 3
ft
diameter
lb
ft
8.98 10 3 ft
2R 1.80 10 2 ft
Problem 1.129
Determine the height that water at 60 °F will rise due to capillary action in a clean,
1
-in.-diameter tube. What will be the height if the diameter is reduced to 0.01 in.?
4
Solution 1.129
θ
2π Rσ
θ
h
γπ R2h
h
2R
The action of surface tension in a tube inserted into a pool is to draw upward (or depress)
2 cos
with respect to the elevation of the surrounding
the liquid in the tube a distance h
R
free surface.
For the specified information:
lb
2 5.03 10 1
cos 0o
ft
h
lb 0.125
62.37 3
ft
12
ft
Similarly, (for R
0.005 in.): h
1.55 10 3 ft
0.186 in.
12 in.
ft
0.125 in.
0.005 in.
0.186 in.
4.65 in.
Problem 1.130
Two vertical, parallel, clean glass plates are spaced a distance of 2 mm apart. If the plates
are placed in water, how high will the water rise between the plates due to capillary action?
Solution 1.130
For equilibrium in the vertical direction,
W
hb 2
cos
h
Thus, (for
h
σℓ
θ
σℓ
2 cos
b
0)
N
2 7.34 10 2
1
m
N
9.80 103 3 0.002 m
m
h
W
7.49 10
3
m 7.49 mm
b
(ℓ ~ width of plates)
Problem 1.131
Walking on water Water striders are insects commonly found on ponds, rivers, and lakes
that appear to “walk” on water. A typical length of a water strider is about 0.4 in., and they
can cover 100 body lengths in one second. It has long been recognized that it is surface tension that keeps the water strider from sinking below the surface. What has been puzzling is
how they propel themselves at such a high speed. They can’t pierce the water surface or
they would sink. A team of mathematicians and engineers from the Massachusetts Institute
of Technology (MIT) applied conventional flow visualization techniques and high-speed
video to examine in detail the movement of the water striders. They found that each stroke
of the insect’s legs creates dimples on the surface with underwater swirling vortices sufficient to propel it forward. It is the rearward motion of the vortices that propels the water
strider forward. To further substantiate their explanation, the MIT team built a working
model of a water strider, called Robostrider, which creates surface ripples and underwater
vortices as it moves across a water surface. Waterborne creatures, such as the water strider,
provide an interesting world dominated by surface tension. (See Problem 1.131.)
(a) The water strider bug shown in the figure below is supported on the surface of a pond
by surface tension acting along the interface between the water and the bug’s legs. Determine the minimum length of this interface needed to support the bug. Assume the bug
weighs 10 4 N and the surface tension force acts vertically upwards. (b) Repeat part (a) if
surface tension were to support a person weighing 750N.
Solution 1.131
σℓ
W  weight
 surface tension
 length of interface
𝒲
For equilibrium, W
W
10 4 N
(a)
N
7.34 10 2
m
(b)
W
750 N
N
7.34 10 2
m
1.36 10 3 m
1000 mm
1m
1.02 10 4 m 6.34 mi !!
1.36 mm
Problem 2.2
The deepest known spot in the oceans is the Challenger Deep in the Mariana Trench of the
Pacific Ocean and is approximately 11,000 m below the surface. Assume that the salt-water
density is constant at 1025kg/m3 and determine the pressure at this depth.
Solution 2.2
GIVEN: Density of the fluid is 1025
kg
m3
, and the depth of the Challenger Deep is 11000 m .
FIND: Pressure at the depth of 11000 m .
SOLUTION:


 1 N ⋅ s 2   1 Pa 
kg   9.807m 

p = ρ gh =  1025 3  
 (11000m )  1 kg ⋅ m   N 

m   s2 


  1
 m2 
= 1.11× 108 Pa
p = 111 MPa gage
Problem 2.3
A closed tank is partially filled with glycerin. If the air pressure in the tank is 6 lb / in.2 and
the depth of glycerin is 10 ft , what is the pressure in lb / ft 2 at the bottom of the tank?
Solution 2.3
2
lb 

 lb   144 in. 
p = γ h + p0 =  78.6 3  (10ft ) +  6 2  

ft 

 in.   1 ft 2 
p = 1,650
lb
ft 2
Problem 2.4
A 3-m- diameter vertical cylindrical tank is filled with water to a depth of 11m . The rest of
the tank is filled with air at atmospheric pressure. What is the absolute pressure at the
bottom of the tank?
Solution 2.4
Known: water filled tank, dia. = 3 m , depth = 11m
Determine: absolute pressure at tank bottom
Strategy: insert information into hydrostatic pressure distribution
Solution:
pbottom = patmos + γ water hbottom
kN 

= 101 kPa +  9.80 3  (11 m ) = 208.80 kPa
m 

pbottom = 209 kPa
Problem 2.5
Blood pressure is usually given as a ratio of the maximum pressure (systolic pressure) to the
minimum pressure (diastolic pressure). Such pressures are commonly measured with a
mercury manometer. A typical value for this ratio for a human would be 120/70, where the
pressures are in mm Hg. (a) What would these pressures be in pascals? (b) If your car tire
was inflated to 120 mm Hg, would it be sufficient for normal driving?
Solution 2.5
p =γh
(a)
(b)
N 

For 120 mm Hg : p =  133 × 103 3  ( 0.120 m ) →
m 

p = 16.0 kPa
N 

For 70 mm Hg : p =  133 × 103 3  ( 0.070 m ) →

m 
p = 9.31 kPa
2

3 N 
−4 lb/in. 
For120 mm Hg : p = 16.0 ×10
 = 2.32 psi
 1.450 ×10
m 2  
N/m 2 

Typical tire pressure is 30 − 35 psi , therefore.
120 mm is insufficent inflation for normal driving
Problem 2.6
An unknown immiscible liquid seeps into the bottom of an open oil tank. Some
measurements indicate that the depth of the unknown liquid is 1.5 m and the depth of the
(
)
oil specific weight = 8.5 kN / m3 floating on top is 5.0 m . A pressure gage connected to
the bottom of the tank reads 65 kPa . What is the specific gravity of the unknown liquid?
Solution 2.6
pbottom = ( γ oil ) ( 5m ) + ( γ u ) (1.5m ) where γ u
γu =
pbottom − ( γ oil ) ( 5 m )
1.5 m
65 ×103
=
N
γu
m3
=
SG =
γ H2O @ 4 C 9.81×103 N
m3
15 × 103
unknown liquid γ
N 
3 N 
−
×
8.5
10

 (5 m )
N
m2 
m3 
= 15 × 103 3
1.5 m
m
→
SG = 1.53
Problem 2.7
A 30-ft- high downspout of a house is clogged at the bottom. Find the pressure at the
bottom if the downspout is filled with 60 °F rainwater.
Solution 2.7
p = ρ gh
Inserting the density of water and the specified column height:
lbm  
ft 

 62.4 3   32.2 2  ( 30ft )
ft  
s 
p=

ft ⋅ lbm 
in2 

32.2
144




lb ⋅ s2  
ft 2 

→
p = 13.0 psig
Problem 2.8
How high a column of SAE 30 oil would be required to give the same pressure as
700 mm Hg?
Solution 2.8
p =γh
For pHg = poil
γ Hg hHg = γ oil hoil
or

kN 

m3  0.700m →
(
)
kN 
 8.95 3 
m 


γ Hg
hoil =
hHg = 
γ oil

133
hoil = 10.4 m
Problem 2.9
Bathyscaphes are capable of submerging to great depths in the ocean. What is the pressure
at a depth of 5 km, assuming that seawater has a constant specific weight of 10.1 kN / m3 ?
Express your answer in pascals and psi.
Solution 2.9
p = γ h + p0
At the surface, p0 = 0 so that
(
)
N 
N

p =  10.1×103 3  5 × 103 m = 50.5 ×106 2

m 
m
lb 

1.450 × 10 −4 2 

N


in. 
p =  50.5 × 106 2  
N
m 


1 2


m


→
p = 50.5MPa
→
p = 7320 psi
Problem 2.10
The deepest known spot in the oceans is the Challenger Deep in the Mariana Trench of the
Pacific Ocean and is approximately 11,000 m below the surface. For a surface density
of 1030 kg/m3 , a constant water temperature, and an isothermal bulk modulus of elasticity
of 2.3 ×109 N/m 2, find the pressure at this depth.
Solution 2.10
GIVEN: Ocean depth of 11,000 m, surface density of 1030 kg/m3, constant water
temperature, and isothermal bulk modulus of elasticity EV , T = 2.3 × 109 N/m 2.
FIND:
Pressure at this depth
SOLUTION:
dp
dp
=
Ev ≡ −
d∀ / ∀ d ρ / ρ
ρ
 p − p0 
p


 ρ  p − p0
dp
 Ev 
ρ
ρ
e
ln
=
→
=
→
=
 
0
 ρ  Ev
Ev
 ρ0 
ρ0
p0
dρ
where: z is positive upward, z = z0 at the surface, and p( z = 0) = p0 , ρ ( z = 0) = ρ0
Substitution into the hydrostatic pressure equation yields:
 p − p0 


dp
E
= −γ = − ρ g = − ρ0 ge v 
dz
p

p0
 p − p0 
 − p − p0  
−



E
E


e  v  dp = − Ev e  v  

p
 − p − p0  
z
  Ev  
= − Ev  e
− 1 = − ρ0 g  dz = − ρ0 gz


0



p0
 p − p0 
−

ρ gz
E
e  v  = 1+ 0
Ev
 p − p0 
 ρ0 gz 
−
 = ln 1 +

Ev 
 Ev 



 1030 kg   9.807 m  ( −11000 m ) 
 ρ gz 
N  


m3  
s2 
p = p0 − Ev ln 1 + 0  = 0 −  2.3 × 109 2  ln 1 + 

Ev 
m  



9 N   1 kg ⋅ m 

×
2.3
10

2 
 1 N ⋅ s2 


m






p = 1.14 ×108
N
m2
→
p = 114 MPa gage
Problem 2.11
A submarine submerges by admitting seawater SG = 1.03 into its ballast tanks. The amount
of water admitted is controlled by air pressure, because seawater will cease to flow into the
tank when the internal pressure (at the hull penetration) is equal to the hydrostatic pressure
at the depth of the submarine. Consider a ballast tank, which can be modeled as a vertical
half-cylinder (R = 8 ft,L = 20ft ) for which the air pressure control valve has failed shut. The
failure occurred at the beginning of a dive from 60 ft to 1000 ft . The tank was initially filled
with seawater to a depth of 2 ft and the air was at a temperature of 40 °F . As the weight of
water in the tank is important in maintaining the boat’s attitude, determine the weight of
water in the tank as a function of depth during the dive. You may assume that tank internal
pressure is always in equilibrium with the ocean’s hydrostatic pressure and that the inlet
pipe to the tank is at the bottom of the tank and penetrates the hull at the “depth” of the
submarine.
Solution 2.11
GIVEN: Ballast tank, L = 20 ft , R = 8ft , initial condition of d = 60 ft, h = 2ft ., Tair = 40 °F,
Final condition d = 1,000 ft.
FIND: Weight of water in ballast tank as a function of depth d during dive.
SOLUTION: The initial water volume in the ballast tank is determined using given
information about the ballast tank,

L  2 −1  2Rh − h2 
2


− ( R − h ) 2Rh − h
Vsw i =
R sin


2
R






2 

20ft 
2
−1  2 ⋅ 8 ⋅ 2 − 2 ft 
2 
=
− ( 8 − 2 ) ft  ⋅ 2 ⋅ 8 ⋅ 2 − 2 ft
(8ft ) sin

 
2 
8ft





b
Vsw
R
h
γ sw
= 145 ft3
The initial air volume is
Vair i =
π R2 L
2
⋅ V wi =
π
2
( 8 ft )2 ( 20 ft ) ⋅145 ft3 = 1866 ft3
During the dive, the ballast tank air pressure is assumed to be in equilibrium with the ocean
hydrostatic pressure. Then
pair = patm + γ sw ( d − h )
Using the ideal gas law,
Mair RTair Mair RTair
=
= patm + γ sw ( h − d )
Vair
Vtank −Vsw
Solving for h ,
h=d+
patm
γ sw
−
M air RTair
γ sw (Vtank −Vsw )
(2)
Vsw is a function of h , given by Eq. (1) for h < R . For h > R , define b = 2R − h and
Vsw =
π R2 L
2

L  2 −1  2Rb − b2 
2


R sin
−
− ( R − b ) 2Rb − b


R
2





(3)
The air mass Mair is
(
lbm 

Mair = ρair iVair i =  0.219 3  1866 ft3
ft 

= 408.7 lbm
)
Where 0.219 lbm/ft3 is the density of air at 40 F and a pressure corresponding to
60 ft of water.
The initial weight of the water is

lb 
Wsw i = γ swVsw i =  62.4 ×1.03 3  145 ft3 = 9320 lb .

ft 

(
)
Pseudocode for procedural language:
hmin = 0
hmax = 2R
h new =
( hmin + hmax )
2
ptank = pair + γ sw h
pocean = patm + γ sw d
ptank < pocean  hmin = h new
TRUE
ptank > pocean  hmax = h new
 (h

−h )
If  max min < errtol 
hmin


(h + h )
h = min max
2
FALSE
Problem 2.12
Determine the pressure at the bottom of an open 5-m- deep tank in which a chemical
process is taking place that causes the density of the liquid in the tank to vary as
 h π
,
 hbot 2 
ρ = ρsurf 1 + sin2 
where h is the distance from the free surface and ρsurf = 1700 kg/m3.
Solution 2.12
GIVEN: H bot = 5 m , ρsurf = 1700
2 h π 
;
=
+
ρ
ρ
1
sin

,
surf
h
2
m3
 bot 
kg
FIND: Pressure at bottom of tank.
SOLUTION:
The pressure gradient is
dp
= ρ =γ
dh
Separating variables, substituting for ρ , and integrating give
p
h
psurf
0
2
h π
 dp = ρsurf g  1 + sin  hbot 2 dh
This integral must be solved numerically.
The Romberg method gives for h = hbot
pbot − psurf = ρsurf g ( 6.080 m )
Setting psurf = 0
N 
m   N ⋅ s2 

pbot = 1700 3   9.807 2  
 ( 6.080 m ) →

m 
s   kg ⋅ m 
pbot = 101×103
N
m2
Problem 2.13
In a certain liquid at rest, measurements of the specific weight at various depths show the
following variation:
h (ft)
(lb/ft3)
0
10
20
30
40
50
60
70
80
90
100
70
76
84
91
97
102
107
110
112
114
115
The depth h = 0 corresponds to a free surface at atmospheric pressure. Determine, through
dp
= −γ , the corresponding variation in pressure and show the
numerical integration of
dz
results on a plot of pressure (in psf) versus depth (in feet).
Solution 2.13
dp
= −γ
dz
Let z = h0 − h (see figure) so
that dz = −dh and therefore
dp = −γ dz = γ dh
h
h0
θ
Thus,
z
pi
hi
0
0
 dp =  γ dh
or
hi
pi =  γ dh
(1)
0
where pi is the pressure at depth hi .
Equation (1) can be integrated numerically using the trapezoidal rule, i.e.,
1 n−1
I =  ( yi + yi +1 )( xi +1 − xi )
2 i =1
where y ~ γ , x ~ h, and n = number of data points.
The tabulated results are given below, along with the corresponding plot of pressure versus
depth.
γ (lb/ft^3)
h (ft)
0
10
20
30
40
50
60
70
80
90
100
Pressure, psf
0
730
1530
2405
3345
4340
5385
6470
7580
8710
9855
70
76
84
91
97
102
107
110
112
114
115
12000
Pressure, p (psf)
10000
8000
6000
4000
2000
0
0
20
40
60
Depth, h (ft)
80
100
Problem 2.15
Under normal conditions, the temperature of the atmosphere decreases with increasing
elevation. In some situations, however, a temperature inversion may exist so that the air
temperature increases with elevation. A series of temperature probes on a mountain give
the elevation–temperature data shown in the table below. If the barometric pressure at the
base of the mountain is 12.1 psia, determine by means of numerical integration the pressure
at the top of the mountain.
Elevation (ft)
Temperature (°F)
5000
5500
6000
6400
7100
7400
8200
8600
9200
9900
50.1 (base)
55.2
60.3
62.6
67.0
68.4
70.0
69.5
68.0
67.1 (top)
Solution 2.15
For an ideal gas, the hydrostatic equation becomes
dp
p
= −γ = − g ρ = − g
dz
RT
then
p2
z
p2
dp
g 2 dz
=
ln
=
−
 p
p1
R z T
p
i
i
In the table below the temperature in °R is given and the integrand
Elevation (ft)
5000
5500
6000
6400
7100
7400
8200
8600
9200
9900
T (°F)
50.1
55.2
60.3
62.6
67.0
68.4
70.0
69.5
68.0
67.1
T (°R)
509.8
514.9
520.0
522.3
526.7
528.1
529.7
529.2
527.7
526.8
1/T (°R)
0.001962
0.001942
0.001923
0.001915
0.001899
0.001894
0.001888
0.00189
0.001895
0.001898
1
tabulated.
T ( °R )
The approximate value of the integral in 
p2 dp
p1
p
= ln
p2
g z2 dz
is 9.34 obtained using
=− 
p1
R z1 T
the trapezoidal rule, i.e.,
I=
1 n−1
1
( yi + yi +1 )( xi +1 − xi ) where y ~ , x ~ elevation,

2 i =1
T
and n = number of data points. Thus,
9900ft  1 
ft
5000ft  T  dz = 9.34 °R
so that (with g = 32.2ft/s2 and R = 1716 ft ⋅ lb/slug ⋅ °R )
ft  
ft 


 32.2 2   9.34
°R 
p
s 
ln p2 = − 
= −0.1753
1
1716 ft ⋅ lb/slug ⋅ °R
(1)
It follows from Eq. (1) with p1 = 12.1psia that
p2 = (12.1psia)e −0.1753 = 10.2 psia
(Note: Since the temperature variation is not very large, it would be expected that the
assumption of a constant temperature would give good results. If the temperature is
assumed to be constant at the base temperature ( 50.1°F ), p2 = 10.1psia , which is only
slightly different from the result given above.)
Problem 2.16
Often young children drink milk ( ρ = 1030 kg / m3 ) through a straw. Determine the
maximum length of a vertical straw that a child can use to empty a milk container,
assuming that the child can develop 75 mm Hg of suction, and use this answer to determine
if you think this is a reasonable estimate of the suction that a child can develop.
Solution 2.16
Known: ρ milk = 1030
kg
m3
, suction = 75 mm Hg
Determine: maximum length of vertical straw, is this reasonable?
Strategy: compute height of equivalent milk column
Solution:
hmax = height of milk column lifted by suction
Δpmax = γ Hg hHg = 75 mm Hg
Δpmax = γ milk hmilk

kg 



m3 
 13600 m3 
Δpmax γ Hg hHg  ρHg 
=
=
hmax,milk =
 ( 75 mm )
 hHg = 
γ milk
γ milk
 1030 kg 
 ρ milk 
hmax,milk = 990.3 mm milk
max. length straw ≈ 1 m
Although this may seem large, adults can routinely lift water much higher through a straw.
Therefore, a 1 m draw seems large, but within reason for a child.
Problem 2.17
(a) Determine the change in hydrostatic pressure in a giraffe’s head as it lowers its head
from eating leaves 6 m above the ground to getting a drink of water at ground level as
shown in the figure below. Assume the specific gravity of blood is SG = 1. (b) Compare the
pressure change calculated in part (a) to the normal 120 mm of mercury pressure in a
human’s heart.
6m
Water
Solution 2.17
(a) For hydrostatic pressure change,
kN 
kN

Δ p = γ h =  9.80 3  ( 6m ) = 58.8 2

m 
m
→
Δ p = 58.8 kPa
(b) To compare with pressure in human heart convert pressure in part (a) to mm Hg:
58.8
kN 

= γ Hg hHg = 133 3  hHg
m
m 

kN
2
hHg = ( 0.442m ) (103
mm
) = 442 mm Hg
m
giraffe hHg = 442 mmHg
human hHg = 120 mmHg
girrafe more than 3.5 times greater
Problem 2.18
What would be the barometric pressure reading, in mm Hg, at an elevation of 4 km in the
U.S. standard atmosphere? Refer to Table C.2 Properties of the U.S. Standard Atmosphere
(SI Units).
Solution 2.18
At an elevation of 4 km,
N
p = 6.166 ×10 4 2 (from the table given in the Problem). Since
m
p =γh
N
m 2 = 0.464 m →
h= =
N
γ
133 ×103 3
m
p
6.166 ×10 4
h = 464 mm
Problem 2.19
Denver, Colorado, is called the “mile-high city” because its state capitol stands on land
1 mile above sea level. Assuming that the Standard Atmosphere exists, what is the pressure
and temperature of the air in Denver? The temperature follows the lapse rate (T = T0 − β z).
Solution 2.19
GIVEN: Denver altitude = 1mile = 5280 ft and standard atmosphere. T = T0 − β z
FIND: Temperature and pressure in Denver.
SOLUTION:
The lapse rate gives:
°R 

T = T0 − β z = 518.67 °R −  0.003566
 ( 5280 ft )
ft 

T = 500 °R = 40 °F
Using Equation and Table:
g
 β z  βR
p = p0 1 −

 T0 
ft 
lb⋅s 2 


 32.2 2  32.2
s 
ft⋅lbm 

  53.35 ft⋅lb  0.003566° R 


 
lbm⋅°R 
ft 
 
°R 
  0.003566 ft  ( 5280 ft )


= (14.696 psia ) 1 − 
518.67°R






p = 12.10 psia
Note: In reasonably good agreement with table in appendix of text.
Problem 2.20
Assume that a person skiing high in the mountains at an altitude of 15,000 ft takes in the
same volume of air with each breath as she does while walking at sea level. Determine the
ratio of the mass of oxygen inhaled for each breath at this high altitude compared to that at
sea level.
Solution 2.20
Let ( )0 denote sea level and ( )15 denote 15,000 ft altitude.
Thus, since m = mass = ρV , where V = volume ,
m0 = ρ0 V0
m15 = ρ15 V15 , where V0 = V15
m15 ρ15 V15 ρ15
=
=
ρ0 V0
ρ0
m0
If it is assumed that the air composition (e.g., % of air that is oxygen) is the same at sea
level as it is at 15,000 ft , use the density values from table of Properties of the U.S.
Standard Atmosphere (BG/EE Units)
ρ0 = 2.377 ×10 −3
slugs
ft
3
and ρ15 = 1.496 ×10 −3
slugs
1.496 ×10 −3 3
m15
ft = 0.629 →
=
slugs
m0 2.377 × 10 −3
ft3
slugs
ft3
so that
m15 = 62.9% of m0
Problem 2.21
Pikes Peak near Denver, Colorado, has an elevation of 14,110 ft. (a) Determine the pressure
at this elevation, based on the equation below. (b) If the air is assumed to have a constant
specific weight of 0.07647 lb/ft3 , what would the pressure be at this altitude?
(c) If the air is assumed to have a constant temperature of 59 °F , what would the pressure
be at this elevation? For all three cases assume standard atmospheric conditions at sea level
as provided in the table of Properties of U.S. Standard Atmosphere at Sea Level).
Solution 2.21
(a)
g
 β z  Rβ
p = pa  1 −

 Ta 
ft
32.174 2
g
s
=
= 5.252
Rβ 
ft ⋅ lb  
°R 
1716 slug ⋅ °R   0.00357 ft 



 
°R 

0.00357
14110ft ) 
(



lb 
ft 


p =  2116.2 2  1 − 
518.67°R
ft  




(b)
(c)
p = pa − γ h = 2116.2
p = pa e
−
→
p = 1240
lb 

0.07647
−

 (14110ft ) →
ft 2 
ft3 
p = 1040
lb
lb
ft 2
lb
ft 2
(abs)
(abs)
gh
RTa
 

ft 
  32.174 2 (14110ft ) 
s



−


ft⋅lb 
lb   1716 slug⋅°R (518.67°R ) 

p =  2116.2 2  e

5.252
ft 
→
p = 1270
lb
ft 2
(abs)
Problem 2.22
g
 β z  Rβ
Equation p = pa  1 −
 provides the relationship between pressure and elevation in the
 Ta 
atmosphere for those regions in which the temperature varies linearly with elevation. Derive
this equation and verify the value of the pressure given in the table of Properties of the U.S.
Standard Atmosphere (SI Units) for an elevation of 5 km.
Solution 2.22
For an ideal gas, the hydrostatic equation becomes
dp
p
= −γ = − g ρ = − g
dz
RT
then
p2 dp
g
z2 dz
p p = − R z T
1
1
Let p1 ~ pa for z1 = 0 , p2 ~ p for z2 = z , and T = Ta − β z.
p dp
g
z
dz
p p = − R 0 Ta − β z
a
z
 βz 

p
g 1
g
g
 ln (Ta − β z ) − ln Ta  =
= −  − ln (Ta − β z )  =
ln
ln 1 −

pa
R β
R β  Ta 
 0 Rβ
g
R
βz  β

p = pa 1 −

 Ta 
For z = 5 km :
m
K
J
pa = 101.33kPa , Ta = 288.15 K , g = 9.807 2 , β = 0.00650 , R = 287
,
kg ⋅ K
m
s
m
9.807 2
s
  287 J   0.0065 K 

  kg⋅K  
m
 
K
3
  0.0065 m  5 × 10 m


p = (101.33kPa ) 1 − 
288.15
K




(
)
pcomputed = 5.40 × 10 4
→
ptabulated = 5.405 × 10
N
m2
4 N
m2
Problem 2.23
–2.5 °C
As shown in the figure below for the U.S. standard atmosphere, the troposphere extends to
an altitude of 11km where the pressure is 22.6 kPa ( abs ). In the next layer, called the
stratosphere, the temperature remains constant at −56.5 °C . Determine the pressure and
density in this layer at an altitude of 15 km. Assume g = 9.77 m / s 2 in your calculations.
Compare your results with those given in Table C.2 Properties of the U.S. Standard
Atmosphere (SI Units).
50
–44.5 °C
47.3 km
(p = 0.1 kPa)
32.2 km (p = 0.9 kPa)
30
20
–56.5 °C
Altitude z (km)
40
Stratosphere
10
20.1 km (p = 5.5 kPa)
11.0 km (p = 22.6 kPa)
p = 101.3 kPa (abs)
15 °C
Troposphere
0
–100
–80
–60
–40
–20
0
+20
Temperature T (°C)
Solution 2.23
For isothermal conditions, for an ideal gas the hydrostatic equation becomes
dp
p
= −γ = − g ρ = − g
dz
RT
then
p2 dp
p
g
z2 dz
p p = ln p21 = − R z T
1
p2 = p1e
− g ( z2 − z1 )
RT0
1
Let
z1 = 11km, p1 = 22.6 kPa, R = 287
m
J
, g = 9.77 2 ,
kg ⋅ K
s
T0 = −56.5 °C + 273.15 = 216.65 K .
p2 = ( 22.6 kPa )


m
3
3
  9.77 2 (15×10 m-11×10 m ) 
s 

− 



J 
287
216.65K
) 


(
kg
K
⋅

 = 12.1 kPa
e  
Using the ideal gas model:
ρ2 =
12.1× 103
N
m2
kg
p
=
= 0.195 3
RT 
J 
m
 287 kg ⋅ K  ( 216.65 K )


In comparison to published values:
pcomputed = 12.1 kPa,
ρ = 0.195
kg
m3
kg
ptabulatted = 12.11 kPa, ρ = 0.1948 3
m
Problem 2.24
The record low sea-level barometric pressure ever recorded is 25.8 in. of mercury. At what
altitude in the standard atmosphere is the pressure equal to this value?
Solution 2.24
For record low pressure,


lb   25.8 in.   ft 2 
lb

p = γ Hg hHg =  847 3  

 = 12.6 2

2


ft   12 in.   144 in. 
in.

ft 

From Table C.1 Properties of the U.S. Standard Atmosphere (BG/EE Units)
@ 0 ft altitude p = 14.696
lb
in.2
@ 5000 ft altitude p = 12.228
lb
in.2
Assume linear variation change in pressure per foot.
14.696
Thus, pressure change per foot =
lb
lb
lb
− 12.228 2
2
in.
in. = 4.936 × 10 −4 in.2
5000 ft
ft
lb 


2 
lb
lb
14.696 2 − a(ft)  4.936 × 10 −4 in.  = 12.6 2
ft 
in.
in.



a=
14.696 − 12.6
4.936 ×10 −4
ft →
a = 4250 ft
Problem 2.25
On a given day, a barometer at the base of the Washington Monument reads 29.97 in. of
mercury. What would the barometer reading be when you carry it up to the observation
deck 500 ft above the base of the monument?
Solution 2.25
Let ( )b and ( )od correspond to the base and observation deck, respectively.
Thus, with H = height of the monument,
pb − pod = γ air H = 7.65 × 10 −2
p = γ Hg h , where γ Hg = 847
lb
ft
lb
ft3
3
lb
(500 ft ) = 38.5 2
ft
and h = barometer reading.
lb
 29.97 
ft  − γ Hg hod = 38.5 2
ft
 12

γ Hg 
lb
38.5 2
29.97


ft × 12 in. = (29.97 − 0.545)in. →
ft  −
hod = 
ft
 12
 847 lb
3
ft
hod = 29.43 in.
Problem 2.26
Aneroid barometers can be used to measure changes in altitude. If a barometer reads 30.1
in. Hg at one elevation, what has been the change in altitude in meters when the barometer
reading is 28.3 in. Hg? Assume a standard atmosphere and that the equation below is
applicable over the range of altitudes of interest.
Solution 2.26
g
R
βz  β

p = pa  1 −

 Ta 
g
β z  Rβ

z1 : p1 = pa 1 −

 Ta 
Rβ
 g
p
→  1
 pa 
= 1−
β z1
Ta
→ z1 =
z2 =
Ta
β
Ta
β
−
−
Rβ
 g
Ta  p1
 
β  pa 
Rβ
 g
Ta  p2
 
β  pa 
Rβ
Rβ 

Ta  p1  g Ta  p2  g 
z2 − z1 =
  −   
β  pa 
β  pa 




For Ta = 288K , β = 0.00650
K
J
m
, Pa = 101kPa, g = 9.81 2 , R = 287
m
kg ⋅ K
s

J 
K
 287 kg ⋅ K   0.00650 m 
Rβ 


=
= 0.190
m
g
9.81 2
s
N 
m


p1 = γ Hg h1 =  133 × 103 3  ( 30.1in.)  2.540 ×10 −2
 = 102 kPa
in. 
m 


N 
m


p2 = γ Hg h2 =  133 × 103 3  ( 28.3 in.)  2.540 ×10 −2
 = 95.6 kPa
in. 


m 
Substitution yields:
 102 kPa 0.190  95.6 kPa 0.190 

 →
z 2 − z1 =
−

K  101kPa 
101kPa



0.00650 

m
288K
z 2 − z1 = 543 m
Problem 2.27
Bourdon gages (see the figure below) are commonly used to measure pressure. When such a
gage is attached to the closed water tank of figure below the gage reads 5 psi. What is the
absolute air pressure in the tank? Assume standard atmospheric pressure of 14.7 psi.
Air
12 in.
Bourdon gage
15
20
10
25
5
30
0
Water
35
6 in.
Solution 2.27
p = γ h + p0
 12 
pgage −  ft  γ H2O = pair
 12 
lb 
 lb
pair =  5 2 + 14.7 2  −
in. 
 in.
(1ft )  62.4 2 

lb
ft 
in.2
144 2
ft
→
p = 19.3 psia
Problem 2.28
On the suction side of a pump, a Bourdon pressure gage reads 40 kPa vacuum. What is the
corresponding absolute pressure if the local atmospheric pressure is 100 kPa (abs)?
Solution 2.28
p ( abs ) = p ( gage ) + p ( atm ) = −40 kPa+100 kPa →
p ( abs ) = 60 kPa
Problem 2.29
A Bourdon pressure gage attached to the outside of a tank containing air reads 77.0 psi
when the local atmospheric pressure is 760 mm Hg. What will be the gage reading if the
atmospheric pressure increases to 773 mm Hg?
Solution 2.29
p ( abs ) = p ( gage ) + p ( atm )
Assuming the absolute pressure of the air in the tank remains constant,
 p ( gage ) + p ( atm )  =  p ( gage ) + p ( atm ) 
i
f
Where i ~ initial state and f ~ final state. Thus,
p f ( gage ) = pi ( gage ) + pi ( atm ) − p f ( atm )
Since,
lb 
ft   1ft 2 


pi ( atm ) = γ Hg hi =  847 3  ( 0.760m )  3.281  
 = 14.7 psia
m   144in.2 
ft 


 773 mm 
p f ( atm ) = 
 (14.7 psia ) = 14.9 psia
 760 mm 
p ( gage ) = 77.0 psi + 14.7 psi − 14.9 psia →
p ( gage ) = 76.8psi
Problem 2.31
A U-tube manometer is used to check the pressure of natural gas entering a furnace. One
side of the manometer is connected to the gas inlet line, and the water level in the other side
open to atmospheric pressure rises 3 in. What is the gage pressure of the natural gas in the
inlet line in in. H2 O and in lb/in2 gage ?
Solution 2.31
patm + ρH2O g Δh = pgas

lbm  
ft  6  
lb ⋅ s2

pgas = 0 +  62.4 3   32.174 2   ft  

ft  
s   12   32.174 ft ⋅ lbm 

pgas = 31.2
lb
ft 2
gage = 6 in. H2 O gage
Problem 2.32
A barometric pressure of 29.4 in. Hg corresponds to what value of atmospheric pressure in
psia, and in pascals?
Solution 2.32
lb  29.4   1ft 2 

ft  
(psi) p = γ h =  847 3  
 →
ft   12   144in.2 

p = 14.4 psia
N 
m


(Pa) p = γ h =  133 × 103 3  ( 29.4in.)  2.540 ×10 −2
 →
in. 
m 


p = 99.3kPa ( abs )
Problem 2.33
For an atmospheric pressure of 101 kPa (abs), determine the heights of the fluid columns in
barometers containing one of the following liquids: (a) mercury, (b) water, and (c) ethyl
alcohol. Calculate the heights including the effect of vapor pressure and compare the results
with those obtained neglecting vapor pressure. Do these results support the widespread use
of mercury for barometers? Why?
Solution 2.33
(Including vapor pressure)
p(atm) = γ h + pv , where pv ~ vapor pressure → h =
(a) hmercury =
(b)
hwater =
(c) halchohol =
p(atm) − pv
γ
N
N
− 1.6 × 10 −1 2
2
m
m = 0.759m
3 N
133 × 10
m3
101×103
101×103 − 1.77 × 103
9.80 × 103
101×103 − 5.9 × 103
7.74 ×103
m = 10.1 m
= 12.3 m
(Without vapor pressure): p( atm ) = γ h → h =
p( atm )
γ
N
m 2 = 0.759m
(a) hmercury =
N
133 × 103 3
m
101× 103
(b)
hwater =
(c) halchohol =
101× 103
9.80 × 103
101× 103
7.74 × 103
m = 10.3 m
m = 13.0 m
hmercury = 0.759 m
hmercury = 0.759 m
hwater = 10.1 m
halchohol = 12.3 m
.vs.
w/vapor pressure
hwater = 10.3 m
halchohol = 13.0 m
wo/vapor pressure
Mercury is a better choice because it requires less height to represent a
comparable pressure difference and the effect of vapor pressure is mush smaller.
Problem 2.34
The closed tank of the figure below is filled with water and is 5 ft long. The pressure gage
on the tank reads 7 psi. Determine: (a) the height, h, in the open water column, (b) the gage
pressure acting on the bottom tank surface AB, and (c) the absolute pressure of the air in
the top of the tank if the local atmospheric pressure is 14.7 psia.
Open
7 psi
Air
h
2 ft
Water
2 ft
A
B
Solution 2.34
p = γ h + p0
(a) p1 = γ H2O (2 ft) + pair
Also p1 = γ H2O h so that
2
lb 
lb 


 lb   144in. 
62.4
h
62.4
2
ft
7
=
+
(
)
 →






ft3 
ft3 


 in.2   ft 2 
h = 18.2 ft
2
2

lb 
 lb   144in.    1ft 
(b) pAB =  62.4 3  ( 4ft ) +  7 2  
 
 →
ft 
 in.   ft 2    144in.2 

(c) pair = 7psi + 14.7 psia= →
pair = 21.7psia
p = 8.73 psi
Problem 2.35
A mercury manometer is connected to a large reservoir of water as shown in the figure
h
below. Determine the ratio, w , of the distances hw and hm indicated in the figure.
hm
Water
hw
hm
hm
Mercury
Solution 2.35
p1 = γ w hw + γ w hm
p1 = p3 = γ m ( 2 hm )
γ w hw + γ w hm = 2γ m hm
hw
Water
γw
(2)
(γ w ) hw = ( 2γ m − γ w ) hm
hm
hm
so that
hw ( 2γ m − γ w )
γ
=
= 2SGm − 1 , where SGm = m = 13.56
γw
hm
γw
Thus,
hw
= 2 (13.56 ) − 1 = 26.1
hm
(1)
(3)
γ ′m
Mercury
Problem 2.36
The U-tube manometer shown in the figure below has two fluids, water and oil ( S = 0.80 ).
Find the height difference between the free water surface and the free oil surface with no
applied pressure difference.
pA
pA
g
Oil
10 cm
Water
Solution 2.36
GIVEN: Soil = 0.8 (see the figure in the problem)
FIND: Free surface height difference.
SOLUTION:
pA + ρ0 gh0 − ρw ghw = pA
ρ 
hw = h0  0  = h0S0
 ρw 
= (10 cm )( 0.8 ) = 8 cm
Δh = h0 − hw
= 10 cm − 8 cm
Δh = 2 cm
Problem 2.37
A U-tube manometer is connected to a closed tank containing air and water as shown in
the figure below. At the closed end of the manometer, the air pressure is 16 psia. Determine
the reading on the pressure gage for a differential reading of 4 ft on the manometer. Express
your answer in psi (gage). Assume standard atmospheric pressure and neglect the weight of
the air columns in the manometer.
Closed valve
Air pressure = 16 psia
4 ft
Air
2 ft
Water
Gage fluid
( γ = 90 lb / ft 3 )
Pressure
gage
Solution 2.37
p1 + γ gf (4 ft) + γ H2O (2 ft) = pgage
lb
lb  
in.2   lb 
lb 


pgage = 16 2 − 14.7 2  144 2  +  90 3  ( 4 ft ) +  62.4 3  ( 2 ft )


in.  
ft   ft 
ft 
 in.

= 627
lb
1ft 2
ft
144in.2
×
2
→
pgage = 4.67 psi
Problem 2.38
The container shown in the figure below holds 60 °F water and 60 °F air as shown. Find
the absolute pressures at locations A, B, and C.
patm = 14.7 lb/in.2
Air
B
10″
8″
14″
A
Water
C
Water
Solution 2.38
GIVEN: Figure, water and air at 60 °F
FIND: Absolute pressures at points A, B, C.
SOLUTION:


γ w =  62.3
lb   ft3 
lb
0.0361
=



ft3   1728 in3 
in3
Modeling the air as an ideal gas:
lb   ft 

14.7 2 

p
lbm
in   12 in 

ρa =
=
= 0.0000442 3
ft ⋅ lb 
RT 
in
 53.35
 ( 520°R )
lbm ⋅ °R 

→ γ a = 0.0000442
The hydrostatic equation gives:
lb 

+  0.0361 3  ( 8 in ) →
in 
in 
pA = 15.0 psia
pB = pA − γ a hB = 15.0
lb 

−  0.0000442 3  (10 in ) →
in 
in 
pB = 15.0 psia
pC = pB + γ w hC = 15.0
lb 

+  0.0361 3  (14 in ) →
in 
in 
pC = 15.5 psia
pA = patm + γ w hA = 14.7
lb
2
lb
2
lb
2
lb
in3
Problem 2.39
A closed cylindrical tank filled with water has a hemispherical dome and is connected to an
inverted piping system as shown in the figure below. The liquid in the top part of the piping
system has a specific gravity of 0.8, and the remaining parts of the system are filled with
water. If the pressure gage reading at A is 60 kPa, determine (a) the pressure in pipe B, and
(b) the pressure head, in millimeters of mercury, at the top of the dome (point C).
Hemispherical dome
SG = 0.8
C
pA =
60 kPa
4m
3m
A
3m
Water
2m
B
Water
Solution 2.39
(
)
(a) pA + ( SG ) γ H2O ( 3m ) + γ H2O ( 2 m ) = pB
N 
N 


pB = 60 kPa+ ( 0.8 )  9.81×103 3  ( 3 m ) +  9.81× 103 3  ( 2 m ) →


m 
m 
pB = 103 kPa
N 
N

(b) pc = pA − γ H2O ( 3 m ) = 60 kPa −  9.81×103 3  ( 3 m ) = 30.6 × 103 2

m 
m
N
30.6 × 103 2
3
pc
m = 0.230 m = 0.230 m  10 mm  → h = 230 mm
h=
=
 m 
γ Hg 133 ×103 N


3
m
Problem 2.40
Two pipes are connected by a manometer as shown in the figure below. Determine the
pressure difference pA − pB , between the pipes.
A
0.5 m
Water
0.6 m
1.3 m
Gage fluid
(SG = 2.6)
Water
B
Solution 2.40
pA + γ H2O ( 0.5m + 0.6m ) − γ gf ( 0.6 m ) + γ H2O (1.3m − 0.5m ) = pB
pA − pB = γ gf ( 0.6 m ) − γ H2O ( 0.5m + 0.6m + 1.3m − 0.5m )
kN 
kN 


= ( 2.6 )  9.81 3  ( 0.6 m ) −  9.81 3  (1.9 m ) →


m 
m 
pA − pB = −3.32 kPa
Problem 2.41
Find the percentage difference in the readings of the two
identical U-tube manometers shown in the figure below.
Manometer 90 uses 90 °C water and manometer 30 uses
h90
h30
30 °C water. Both have the same applied pressure
g
difference. Does this percentage change with the magnitude
of the applied pressure difference? Can the difference
between the two readings be ignored?
Manometer 90
Manometer 30
Solution 2.41
GIVEN: Figure, Two identical U-tube manometers. Manometer 90 uses 90°C water
while manometer 30 uses 30°C water. Same pressure difference applied across each
manometer.
FIND: Percent difference in readings. Does this percent difference change with the applied
pressure difference? Can the difference in the two manometer readings be ignored?
SOLUTION:
Apply the manometer rule:
pB = pA + ρw gh → h90 =
pB − pA
ρ90w g
h30 =
pB − pA
ρ30w g
Using the 30 °C water as a reference
1 
 1
−

 h90 − h30 
ρ

ρ90w ρ30w 

 ×100 =  30w − 1 × 100
×
=
100


1


 h30 
 ρ90w 


ρ30w
 h90 − h30 
h −h 
 996 
− 1 × 100 →  90 30  = 3.2%

 ×100 = 
 965 
 h30 
 h30 
Note that this percent difference does not change with the applied pressure difference and
the difference in the two manometer readings cannot be ignored in most cases.
Problem 2.42
A U-tube manometer is connected to a closed tank as
shown in the figure below. The air pressure in the
tank is 0.50 psi and the liquid in the tank is oil
lb
( γ = 54.0 3 ). The pressure at point A is 2.00 psi.
ft
Determine: (a) the depth of oil, z , and (b) the
differential reading, h, on the manometer.
Open
Air
z
Oil
h
A
2 ft
SG = 3.05
Solution 2.42
(a)
pA = γ oil z + pair
lb   144 in.2 
 lb
−
2
0.5



2
in.2   ft 2 
pA − pair  in.
=
z=
lb
γ oil
54.0 3
ft
(a )
(b)
(
→
z = 4.00 ft
)
pA + γ oil ( 2 ft ) − ( SG ) γ H2O h = 0
2
lb 
 lb   144 in.  
 +  54.0 3  ( 2 ft )
 2 2  
2
ft 
p + γ (2 ft)  in.   ft
 
h = A oil
=
lb
( SG ) γ H2O
(3.05 )  62.4 3 
ft 

(
)
→
h = 2.08 ft
Problem 2.43
For the inclined-tube manometer of the figure below, the pressure in pipe A is 0.6 psi. The
fluid in both pipes A and B is water, and the gage fluid in the manometer has a specific
gravity of 2.6. What is the pressure in pipe B corresponding to the differential reading
shown?
Water
B
Water
A
3 in.
8 in.
SG = 2.6
3 in.
30°
Solution 2.43
 3 
 8 
 3 
pA + γ H2O  ft  − γ gf  ft  sin30° − γ H2O  ft  = pB
 12 
 12 
 12 
 8 
pB = pA − γ gf  ft  sin30°
 12 
lb  
in.2 
lb   8 


=  0.6 2   144 2  − ( 2.6 )  62.4 3   ft  ( 0.5 )


in.  
ft 
ft   12 


= 32.3
lb
1 ft 2
×
ft 2 144 ft 2
→
pB = 0.224 psi
Problem 2.44
A flowrate measuring device is installed in a horizontal pipe through which water is
flowing. A U-tube manometer is connected to the pipe through pressure taps located 3 in.
on either side of the device. The gage fluid in the manometer has a specific weight of
lb
122 3 . Determine the differential reading of the manometer corresponding to a pressure
ft
lb
drop between the taps of 0.5 2 .
in.
Solution 2.44
Let p1 and p2 be pressures at pressure taps.
Apply manometer equation between
p1 and p2.
Flowmeter
Flow
p1
(Water)
γH O
2
p1 + γ H2O ( h1 + h ) − γ gf h − γ H2O h1 = p2
p2
h1
h
γ gf
lb  
in.2 

 0.5 2   144 2 
in.  
ft 

p1 − p2
=
h=
lb
lb
γ gf − γ H2O
122 3 − 62.4 3
ft
ft
→
h = 1.45 ft
Problem 2.45
The sensitivity Sen of the micromanometer shown in the figure
below is defined as
Sen =
H
.
pL − pR
pR
pL
h
ρA
ρA
Find the sensitivity of the micromanometer in terms of the
densities ρ A and ρ B . How can the sensitivity be increased?
H
ρB
Solution 2.45
GIVEN: Figure and sensitivity defined as: Sen =
H
.
pL − pR
FIND: Sensitivity as a function of fluid densities. How can the sensitivity increase?
SOLUTION:
Apply manometer rule,
pR + γ Ah + ( γ B − γ A ) H = pL
pL − pR = γ Ah + ( γ B − γ A ) H
Sen =
H
H
=
PL − PR γ Ah + ( γ B − γ A ) H
→
Sen =
1
 h 
 + (γ B − γ A )
H 
γA 
The sensitivity can be increased by decreasing the denominator.
➔ decrease density difference or
h
➔ decrease
by increasing the ratio of reservoir area to tube area.
H
Problem 2.46
The cylindrical tank with hemispherical ends shown
in the figure below contains a volatile liquid and its
kg
vapor. The liquid density is 800 3 , and its vapor
m
density is negligible. The pressure in the vapor is
120 kPa ( abs ) and the atmospheric pressure is
Open
Vapor
1m
1m
Liquid
h
1m
101 kPa ( abs ). Determine: (a) the gage pressure
reading on the pressure gage, and (b) the height, h,
of the mercury, in the manometer.
Mercury
Solution 2.46
kg  
m
N

(a) Let γ = specific weight of liquid =  800 3   9.81 2  = 7850 3

m 
s 
m
pvapor (gage) = 120 kPa(abs) − 101kPa ( abs ) = 19 kPa
pgage = pvapor + γ ( l m ) = 19 × 103
N 

+  7850 3  ( l m ) →
m 
m 
N
2
pgage = 26.9 kPa
(b) pvapor (gage) + γ ( lm ) − γ Hg ( h ) = 0
19 × 103
N 
N 


+  7850 3  ( l m ) − 133 ×103 3  ( h ) = 0 →

m 
m 
m 
N
2
h = 0.202 m
Problem 2.47
Determine the elevation difference, Δh, between the water levels in the two open tanks
shown in the figure below.
SG = 0.90
0.4 m
Δh
1m
Water
Solution 2.47
Let subscript 1 indicate the surface of the left tank, and subscrip 2 the surface of the right
tank.
p1 − γ H2O h + ( SG ) γ H2O ( 0.4 m ) + γ H2O ( h − 0.4 m ) + γ H2O ( Δh ) = p2
Δh = 0.4 m − ( 0.9 )( 0.4 m ) →
Δh == 0.040 m
Problem 2.48
What is the specific gravity of the liquid in the left leg of the U-tube manometer shown in
the figure below?
patm
patm
g
Unknown
fluid
15 cm
20 cm
Water
(S = 1)
10 cm
Solution 2.48
GIVEN: Figure
FIND: Specific gravity SG of unknown fluid
SOLUTION:
h10 = 10cm

Let h15 = 15cm
h = 20cm
 20
Apply manometer rule,
patm + ρw g ( h20 − h10 ) − ρu gh15 = patm
SG =
ρu h20 − h10 20cm − 10cm
=
=
→
ρw
h15
15cm
SG = 0.667
Problem 2.49
For the configuration shown in the figure below what must be the value of the specific
lb
weight of the unknown fluid? Express your answer in 3 .
ft
Open
Open
Water
5.5 in.
4.9 in.
Unknown
fluid
3.3 in.
1.4 in.
Solution 2.49
Let γ be specific weight of unknown fluid.
Applying the manometer rule:
 ( 5.5 − 1.4 ) 
 ( 3.3 − 1.4 ) 
 ( 4.9 − 3.3 ) 
ft  − γ 
patm + γ H 2O 
ft  = patm
ft  − γ H2O 
12
12
12






γ=
γ H O ( 5.5 − 1.4 ) − ( 4.9 − 3.3 )  in. 
2
( 3.3 − 1.4 ) in.
lb   4.1 − 1.6 
=  62.4 3  
 →
ft   1.9 

γ = 82.1
lb
ft3
Problem 2.50
The manometer shown in the figure below has an air bubble either in (a) the right
horizontal line or (b) the left vertical leg. Find h1 − h2 for both cases if pA = pB .
pA
pA = pB = patm = 14.7 psia
Water
pB
Water
p = 16.1 psia
h1
h2
h
Air bubble
12″
Water
Air bubble
Solution 2.50
GIVEN: Figure, manometer with small pockets of air.
FIND: h1 = h2 if (a) air pocket in horizontal line and (b) air pocket in vertical line.
SOLUTION:
(a) Air pocket in horizontal line. Apply the manometer rule between the left liquid
surface (A) and the right liquid surface (B),
pA = pB + γ h2 − γ h1
h2 − h1 =
1
γ
( pA − pB )
pA = pB = 14.7 psia
h1 = h2
(b) Air pocket in (left) vertical line. Apply the manometer rule
pA = pB + γ h2 − γ ( h1 − h )
pA = pB = 14.7 psia
h1 − h2 = h
NOTE: For above analyses, the hydrostatic pressure of the air pocket has been neglected.
Problem 2.51
The U-tube manometer shown in the figure below has legs that are
1.00 m long. When no pressure difference is applied across the
manometer, each leg has 0.40 m of mercury. What is the maximum
pressure difference that can be indicated by the manometer?
Water
H
1.00 m
Hg
Solution 2.51
0.2 m
GIVEN: Manometer in the figure in the problem. With no pressure difference applied
across manometer, each mercury leg is 0.40 m high.
FIND: Maximum pressure difference that can be indicated by the manometer.
SOLUTION:
The maximum pressure difference that can be indicated is illustrated by the
sketch on the right. Applying the manometer rule,
p1
p2
p1 = p2 + γ Hg H − γ w H
(
)
p1 − p2 = (γ Hg − γ w )H = ρHg − ρw gH
Using table in the Appendix, and assuming 20 º C fluid
p1 − p2 = (13550 − 998)
= 123100
 N ⋅ s2 
kg 
m
9.807
1.0m
) 


(
m3 
s2 
 kg ⋅ m 
N
m2
p1 − p2 = 123.1 kPa
H
Problem 2.52
Both ends of the U-tube mercury manometer of the figure below are
initially open to the atmosphere and under standard atmospheric
pressure. When the valve at the top of the right leg is open, the level
of mercury below the valve is h1 . After the valve is closed, air
pg
Valve
hi
pressure is applied to the left leg. Determine the relationship between
the differential reading on the manometer and the applied gage
pressure, pg . Show on a plot how the differential reading varies with
Mercury
pg for h1 = 25, 50, 75, and 100 mm over the range
0 ≤ pg ≤ 300 kPa . Assume that the temperature of the trapped air remains constant.
Solution 2.52
With the valve closed and a pressure, pg , applied,
pg
pa
pg − γ Hg Δh = pa
Δh =
pg − pa
γ Hg
Valve
hi
Δh
(1)
Mercury
Where pg and pa are gage pressures.
For isothermal compression of trapped air
p
ρ
= Constant →
pi V i = pf V f
where V is air volume, p is absolute pressure, i & f refer to initial and final states.,
respectively.
patm V i = ( pa − patm ) V f
(2)
For air trapped in right leg, V i = hi (Area of tube) so that Eq. (2) can be written as


 hi

pa = patm 
− 1
Δh
 hi −

2


(3)




hi  
1 
Substitute Eq.(3) into Eq.(1) to obtain: Δh =
 p + p 1 −

γ Hg  g atm  h − Δh  
i

2  

pg + patm 
2 pg hi
2 
=0
Eq. (4) can be expressed in the form: ( Δh ) −  2 hi +
 Δh +

2γ Hg 
γ Hg

(4)
The roots of this quadratic equation are

pg + patm 
Δh =  hi +
±

2γ Hg 

2

pg + patm  2 pg hi
 hi +
 −

γ Hg
2γ Hg 

(5)
To evaluate Δh , the negative sign is used since Δh = 0 for pg = 0 .
Tabulated values of Δh for various values of pg are given in the following table for different
values of hi (with patm = 101kPa and γ Hg = 133kN/m3 ). A plot of the data follows.
hi (m)
patm
(kPa)
hg
(kN/m3)
pg
(kPa)
Δh
(hi = 0) (m)
Δh
(hi=0.025) (m)
Δh
(hi=0.05) (m)
Δh
(hi=0.075) (m)
Δh
(hi=0.1) (m)
0.025
101
133
0
0
0
0
0
0
0.05
101
133
30
0
0.0110
0.0212
0.0306
0.0394
0.075
101
133
60
0
0.0182
0.0354
0.0517
0.0672
0.1
101
133
90
0
0.0231
0.0454
0.0668
0.0874
101
133
120
0
0.0268
0.0528
0.0781
0.1026
101
133
150
0
0.0296
0.0585
0.0867
0.1143
101
133
180
0
0.0318
0.0630
0.0936
0.1236
101
133
210
0
0.0335
0.0666
0.0991
0.1312
101
133
240
0
0.0350
0.0696
0.1037
0.1374
101
133
270
0
0.0362
0.0721
0.1075
0.1426
101
133
300
0
0.0372
0.0742
0.1108
0.1470
0.16
hi = 0.10
0.14
0.12
hi = 0.075
Δh, m
0.1
0.08
hi = 0.050
0.06
0.04
hi = 0.025
0.02
hi = 0
0
0
50
100
150
200
pg, kPa
250
300
350
Problem 2.53
The inverted U-tube manometer of the figure below
contains oil ( SG = 0.9 ) and water as shown. The pressure
differential between pipes A and B, pA − pB, is −5 kPa .
Determine the differential reading h.
Oil
h
0.2 m
A
0.3 m
Water
B
Solution 2.53
pA − γ H2O ( 0.2 m ) + γ oil ( h ) + γ H2O ( 0.3 m ) = pB
h=
( pB − pA ) + γ H O ( 0.2 m ) − γ H O ( 0.3 m )
2
5 × 103
=
2
γ oil
N 
N 
− 9.80 × 103 3  ( 0.1m )
2 
m 
m 
N
8.95 ×103 3
m
→
h = 0.449 m
Problem 2.54
An inverted U-tube manometer containing oil ( SG = 0.8 ) is located between two
reservoirs as shown in the figure below. The reservoir on the left, which contains carbon
tetrachloride, is closed and pressurized to 8 psi . The reservoir on the right contains water
and is open to the atmosphere. With the given data, determine the depth of water, h, in the
right reservoir.
Carbon tetrachloride
8 psi
0.7 ft
3 ft
Oil
Water
1 ft
h
1 ft
Solution 2.54
Let pA be the air pressure in left reserviour. Manometer equation can be written as
pA + γ CCl4 ( 3ft − 1ft − 1ft − 0.7 ft ) + γ oil ( 0.7 ft ) − γ H2O ( h − 1ft − 1ft ) = 0
h=
pA + γ CCl4 ( 0.3ft ) + γ oil ( 0.7ft )
γH O
+ 2ft
2
in.2  
lb 
lb 
 lb  

8
144
+  99.5 3  ( 0.3ft ) +  57.0 3  ( 0.7 ft )

2 
2 


ft  
ft 
ft 
 in.  

+ 2 ft
=
lb
62.4 3
ft
h = 21 ft
Problem 2.55
The sensitivity Sen of the manometer shown in the figure
h
.
below can be defined as: Sen =
pL − pR
pL
pR
H
HL
Three manometer fluids with the listed specific gravities S
are available:
Fluid
hL
Kerosene, SG = 0.82 ;
SAE 10 oil , SG = 0.87 ; and
HR
hR
h
Water
Water
γW
γW
AL
AR
Normal octane, SG = 0.71 .
Which fluid gives the highest sensitivity? The areas AR and
AL are much larger than the cross-sectional area of the
manometer tube, so H << h .
Solution 2.55
GIVEN: The figure in the problem, three manometer fluids, kerosene ( SG = 0.82 ),
SAE 10 oil ( SG = 0.87 ), and normal octane ( SG = 0.71 ).H << h .
FIND: Manometer fluid that gives highest sensitivity.
SOLUTION:
Apply manometer rule,
pR + γ w ( H R + hR ) − γ f ( hR ) + γ f ( hL ) − γ w ( H L + hL ) = pL
pR + γ w ( H R + hR − H L − hL ) + γ f ( hL − hR ) = PL
pR + γ w ( H R − H L + hR − hL ) + γ f ( h ) = pL
pR − γ w ( H L − H R + hL − hR ) + γ f ( h ) = pL
pR − γ w ( H + h ) + γ f ( h ) = pL
pR − pL = γ w ( H + h ) + γ f ( h )
(
)
H << h → pR − pL ≈ γ w − γ f h
Sen =
1
h
h
≈
=
pR − pL
γ w − γ f h ρw g SGw − SG f
(
)
(
)
Sensitivity maximized for SG f closest to SGw = 1 → SAE 10 oil.
Sen =
1
kg  
m

1000 3  9.81 2  (1 − 0.87 )
m 
s 

Sen = 0.000784
m
Pa
→
=
kg ⋅ m
1
m 3 s2
N
×
×
( 9810 )(1 − 0.87 ) kg m Pa ⋅ m 2 N ⋅ s2
Sen = 0.784
mm
Pa
Problem 2.56
In the figure below pipe A contains gasoline ( SG = 0.7 ),
pipe B contains oil ( SG = 0.9 ), and the manometer fluid
is mercury. Determine the new differential reading if the
pressure in pipe A is decreased 25 kPa, and the pressure in
pipe B remains constant. The initial differential reading is
0.30 m as shown.
B
0.4 m
Oil
A
0.3 m
Mercury
Gasoline
Solution 2.56
For the initial configuration:
B
pA + γ gas ( 0.3 m ) − γ Hg ( 0.3 m ) − γ oil ( 0.4 m ) = pB (1)
With a decrease in pA to p ' A gage fluid levels change
0.4 m
A
a
as shown on figure.
0.3 m
Gasoline
Thus, for final configuration:
′ + γ gas ( 0.3 − a ) − γ Hg ( Δh ) − γ oil ( 0.4 + a ) = pB
pA
Oil
Mercury
a
(2)
Where all lengths are in m. Subtract Eq. (2) from Eq. (1) to obtain,
′ + γ gas ( a ) − γ Hg ( 0.3 − Δh ) + γ oil ( a ) = 0
pA − pA
Since 2 a + Δh = 0.3 (see figure) then, a =
(3)
0.3 − Δh
2
and from Eq. (3),
0.3 − Δh 
 0.3 − Δh 
′ + γ gas 
pA − pA
 − γ Hg ( 0.3 − Δh ) + γ oil 
=0
2
2




′ + γ gas ( 0.15 ) − γ Hg ( 0.3 ) + γ oil ( 0.15 )
pA − pA
Δh =
γ gas γ oil
−γ Hg +
+
2
2
for pA − p ' A = 25 kPa
25
Δh =
kN
kN 
kN 
kN 



+ (0.7)  9.81 3  ( 0.15 m ) − 133 3  ( 0.3 m ) + ( 0.9 )  9.81 3  ( 0.15 m )
2
m

m 

m 

m 
kN (0.7) 
kN  (0.9) 
kN 
−133 3 +
 9.81 3  +
 9.81 3 
2 
2 
m
m 
m 
Δh = 0.100 m
Δh
Problem 2.57
The mercury manometer of the figure below indicates a differential reading of 0.30 m when
the pressure in pipe A is 30-mm Hg vacuum. Determine the pressure in pipe B.
Water
0.50 m
Oil
A
0.15 m
Mercury
B
0.30 m
Solution 2.57
pB + γ oil ( 0.15 m+0.30 m ) − γ Hg ( 0.3 m ) − γ H2O ( 0.15 m ) = pA
where pA = −γ Hg ( 0.030 m )
Thus,
pB = −γ Hg ( 0.030 m ) − γ oil ( 0.45 m ) + γ Hg ( 0.3 m ) + γ H2O ( 0.15 m )
kN 
kN 
kN 
kN 




= − 133 3  ( 0.030 m ) −  8.95 3  ( 0.45 m ) + 133 3  ( 0.3 m ) +  9.80 3  ( 0.15 m )

m 

m 

m 

m 
pB = 33.4 kPa
Problem 2.58
Consider the cistern manometer shown in the figure below. The scale is set up on the basis
that the cistern area A1 is infinite. However, A1 is actually 50 times the internal crosssectional area A 2 of the inclined tube. Find the percentage error (based on the scale
reading) involved in using this scale.
pL
A2
pH
x
Air
0
Air
Oil, S = 0.85
θ = 30°
A1
Solution 2.58
GIVEN: The figure in the problem with A1= 50 A 2.
pH
FIND: Percent error in using a scale based on A1 as
pL
infinite.
SOLUTION:
Apply manometer rule, using the elevation changes
shown in the sketch.
0
ΔhT
Δhc
pH − pL = γ ( Δhc + ΔhT )
pH − pL
γ
= Δhc + ΔhT
(1)
Δhc and ΔhT are vertical drop & rise of fluid level from the scale’s zero.
Conservation of mass requires,  Δhc A1 = xA2 or
Geometry 
ΔhT = x sin30°
Eqs. (2) and (3) into Eq. (1)
pH − pL
γ
=
x=
Δhc =
xA2
A1
A 1
xA2
+ x sin30° = x  2 + 
A1
 A1 2 
pH − pL
A 1
γ 2+ 
 A1 2 
(2)
(3)
For an infinite cistern area, A1 = ∞ 
x∞ =
pH − pL
2( pH − pL )
=
γ
A 1
γ 2+ 
 A1 2 
  A2 1  
 x −x
 x∞ 
100
1
100
=
−
=
+  − 1 100
Percent error = % E =  ∞
2 



 x 
 x

  A1 2  
For
A2
1
  1 1 
=  % E =  2  +  − 1 100
A1 50
  50 2  
→
% E = 4%
Problem 2.59
The cistern shown in the figure below has a diameter D that is 4 times the diameter d of
the inclined tube. Find the drop in the fluid level in the cistern and the pressure difference
( pA − pB ) if the liquid in the inclined tube rises l = 20 in. The angle θ is 20° . The fluid’s
specific gravity is 0.85 .
pA
d
pB
ℓ
0
θ
D
Solution 2.59
GIVEN: The figure in the problem, D = 4d , l = 20 in. , θ = 20º, S = 0.85.
FIND: pA − pB
SOLUTION:
Conservation of mass requires that cistern level
drops as the tube level rises, as show in the
sketch.
hT = vertical rise in tube, hc = drop in cistern
fluid level.
pA
pB
0
hT
hc
2–36
Apply manometer rule,
pB = pA − γ ( hT + hc )
Conservation of mass requires,
hc Ac = lAT
 AT   d 2
hc = l 
=l 
 AC   D 
hT = l sinθ ,
2

d  
pA − pB = γ ( hT + hc ) = γ l sin θ +   
 D  

2
3


lb 

 1    ft
pA − pB =  62.4 3  ( 0.85 )( 20 in.) sin 20° +    
 →
ft 
 4    1728 in3 


pA − pB = 0.248 psia
Problem 2.60
The inclined differential manometer of the figure below
contains carbon tetrachloride. Initially the pressure
differential between pipes A and B, which contain a brine
( SG = 1.1 ), is zero as illustrated in the figure. It is
desired that the manometer give a differential reading of
12 in. (measured along the inclined tube) for a pressure
differential of 0.1 psi . Determine the required angle of
inclination, θ .
A
B
Brine
Brine
12 in.
Carbon
tetrachloride
θ
Solution 2.60
′ − pB′ the left column falls a distance, a , and the right
When pA − pB is increased to pA
column rises a distance b along the inclined tube as shown in the figure. For this final
configuration:
′ + γ br ( hi + a ) − γ CCl ( a + b sin θ ) − γ br ( hi − b sin θ ) = pB′
pA
4
(
)
′ − pB′ + γ br − γ CCl ( a + b sin θ ) = 0
pA
4
(1)
The differential reading, Δ h, along the tube is
Δh =
A
B
a
+b
sin θ
Brine
hi
(
)
′ − pB′ + γ br − γ CCl ( Δ h sin θ ) = 0
pA
4
sin θ =
′ − pB′ )
− ( pA
(γ br − γ CCl ) ( Δh )
4
′ − pB′ = 0.1psi
For pA
lb  
in.2 

−  0.1 2  144 2 
in.  
ft 

sin θ =
= 0.466

lb 
lb   12 

(1.1)  62.4 3  − 99.5 3   12 ft 
ft 
ft  



For Δh = 12 in.,
b
Brine
Thus, from Eq. (1)
θ = 27.8°
a
Carbon
tetrachloride
θ
Problem 2.61
Determine the new differential reading
along the inclined leg of the mercury
manometer of the figure below, if the
pressure in pipe A is decreased 10 kPa and
the pressure in pipe B remains unchanged.
The fluid in A has a specific gravity of 0.9
and the fluid in B is water.
A
SG = 0.9
Water
100 mm
30°
B
80 mm
50 mm
Mercury
Solution 2.61
For the initial configuration:
pA + γ A ( 0.1) + γ Hg ( 0.05sin30° ) − γ H2O ( 0.08 ) = pB
where all length are in m. When pA
decreases, left column moves up a
distance, a , and right column moves
down a distance, a , as shown in the
figure.
A
(1)
SG = 0.9
Water
100 mm
30°
B
80 mm
a
50 mm
For the final configuration:
′ + γ A ( 0.1 − a sin30° ) + γ Hg ( a sin30° + 0.05sin30° + a ) − γ H O ( 0.08 + a ) = pB
pA
2
′ is the new pressure in pipe A. Subtract Eq. (2) from Eq. (1) to obtain
Where pA
′ − pB′ + γ A ( a sin30° ) − γ Hg a ( sin30° + 1) + γ H O ( a ) = 0
pA
2
Thus, a =
′ )
− ( pA − pA
γ A sin30° − γ Hg ( sin30° + 1) + γ H2O
′ = 10 kPa
For pA − pA
kN
m2
a=
= 0.0540 m
kN 
kN 
kN


( 0.9 )  9.81 3  ( 0.5 ) − 133 3  ( 0.5 + 1) + 9.81 3
m 
m 
m


−10
New differential reading, Δ h, measured along inclined tube is equal to
Δh =
a
0.0540 m
+ 0.05 + a =
+ 0.05 m+0.0540 m = 0.0212 m
sin30°
0.5
a
Mercury
(2)
Problem 2.62
A student needs to measure the air pressure inside a compressed air tank but does not have
ready access to a pressure gage. Using materials already in the lab, she builds a U-tube
manometer using two clear 3-ft- long plastic tubes, flexible hoses, and a tape measure. The
only readily available liquids are water from a tap and a bottle of corn syrup. She selects the
corn syrup because it has a larger density ( SG = 1.4 ). What is the maximum air pressure, in
psia, that can be measured?
Solution 2.62
Known: two 3-ft- long clear tubes, unknown length flexible hose, tape measure,
corn syrup ( SG = 1.4 )
Determine: Maximum compressed air pressure
Strategy: reflect on possible physical considerations, apply hydrostatic pressure
equation
Solution:
Form u-tube manometer by connecting bottom of tubes with hose; top of one
tube is connected to the tank.
Assume that tubes are of equal diameter.
Assume that hose is not transparent
➔ opaque hose cannot contribute to usable manometer “height”
Set tubes at equal elevation.
Fill tubes to bottom of tube with corn syrup
➔ maximum difference that can be observed is 3 ft
Δp = γΔh = SGγ H2O Δh
 1 ft 2 
lb 

= (1.4 )  62.4 3  ( 3 ft ) 
= 1.82 psig=16.52 psia
 144 in2 
ft 



Δpmax = 16.5 psi
Problem 2.63
Determine the ratio of areas,
A1
, of the two
A2
A
manometer legs of the figure below if a change in
pressure in pipe B of 0.5 psi gives a corresponding
change of 1 in. in the level of the mercury in the
right leg. The pressure in pipe A does not change.
B
Oil
(SG = 0.8)
Water
Area = A2
Area = A1
Mercury
Solution 2.63
For the initial configuration (see the figure):
pA + γ H2O ( hi + Δ hi ) − γ Hg ( Δ hi ) − γ oil ( hi ) = pB
A
B
(1)
hi
When pB increases the right column falls a
distance, a , and the left column rises a distance,b.
Since the volume of the liquid must remain constant,
A a
A1 b = A2 a or 1 = .
A2 b
Oil
(SG = 0.8)
Water
a
Δhi
Area = A1
b
Mercury
For the final configuration, with pressure in B equal
to pB′ :
pA + γ H2O ( hi + Δ hi − b ) − γ Hg ( Δ hi − a − b ) − γ oil ( hi + a ) = pB′
(2)
Subtract Eq. (1) from Eq. (2) to obtain
−γ H2O ( b ) + γ Hg ( a + b ) − γ oil ( a ) = pB′ − pB
b=
Area = A2
( pB′ − pB ) − γ Hg ( a ) + γ oil ( a )
γ Hg − γ H O
2
pB′ − pB = 0.5 psi and a = 1in.
lb  
in.2  
lb   1 
lb   1 


−  847 3   ft  + ( 0.8 )  62.4 3   ft 
0.5
144


2 
2 

in.  
ft  
ft   12 
ft   12 


b=
= 0.00711ft
lb
lb
847 3 − 62.4 3
ft
ft
1
ft
A1 a
12
= =
A 2 b 0,00711ft
→
A1
= 11.7
A2
Problem 2.64
Determine the change in the elevation of the
mercury in the left leg of the manometer of
the figure below as a result of an increase in
pressure of 5 psi in pipe A while the pressure
in pipe B remains constant.
Water
Oil (SG = 0.9)
A
18 in.
B
6 in.
30°
1
_ -in.2
diameter
Solution 2.64
12 in.
1
_ -in.-diameter
4
Mercury
For the initial configuration:
 18 
6

 12 
pA + γ H2O   − γ Hg  sin30°  − γ oil   = pB
 12 
 12

 12 
(1)
Where all lengths are in ft. When pA
Water
Oil (SG = 0.9)
A
′ the left column falls by
increases to pA
the distance, a , and the right column
moves up the dance, b, as shown in the
figure.
18 in.
B
6 in.
Area = A1
b
a
30°
1
_ in.
2
diameter
1
_ in. diameter
4
Area = A2
Mercury
For the final configuration:
18
6
12
′ + γ H O  + a  − γ Hg  a + sin30° + b sin30°  − γ oil  − b sin30°  = pB (2)
pA
2
12
 12



 12

Subtract Eq. (1) from Eq. (2) to obtain
′ − pA + γ H O ( a ) − γ Hg ( a + b sin30° ) + γ oil ( b sin30° ) = 0
pA
2
(3)
The volume of liquid must be constant: A1 a = A2 b,
2
2
1 
1 
 in.  a =  in.  b ➔ b = 4 a
2 
4 
Thus, Eq. (3) can be written as
′ − pA + γ H O ( a ) − γ Hg ( a + 4 a sin30° ) + γ oil ( 4 a sin30° ) = 0
pA
2
in.2 
 lb  
−  5 2   144 2 
′ − pA )
− ( pA
ft 
 in.  
a=
=
lb 
lb 
lb 
γ H2O − γ Hg (3) + γ oil (2)

62.4 3 −  847 3  (3) + (0.9)  62.4 3  (2)
ft 
ft 
ft 

a = 0.304 ft down
12 in.
Problem 2.65
The U-shaped tube shown in the figure below initially contains
water only. A second liquid with specific weight, γ , less than
water is placed on top of the water with no mixing occurring.
Can the height, h, of the second liquid be adjusted so that the
left and right levels are at the same height? Provide proof of
your answer.
γ
h
D1 = 1.5 D2
Water
Solution 2.65
The pressure at point (1) must be equal to the pressure at point
(2) since the pressures at equal elevations in a continuous mass
of fluid must be the same.
p1 = γ h
D2
γ
h
D1 = 1.5 D2
Water
p2 = γ H2O h
There two pressures can only be equal if γ = γ H2O .
Since γ ≠ γ H2O , the configuration shown in the figure is not possible.
No
D2
Problem 2.66
An inverted hollow cylinder is pushed into the water as is
shown in the figure below. Determine the distance, , that the
water rises in the cylinder as a function of the depth, d , of the
lower edge of the cylinder. Plot the results for0 ≤ d ≤ H ,
when H is equal to 1 m. Assume the temperature of the air
within the cylinder remains constant.
Water
H
d
ℓ
D
Open end
Solution 2.66
For constant temperature compression within the cylinder, p i V i = p f V f (1)
where V is the air volume, and i and f refer to the initial and final states, respectively
p f = γ ( d − ) + patm
pi = patm
Vi=
π
4
D2 H
V f =
π
4
D2 ( H − )
π
π

From Eq. (1): patm  D 2 H  = ( γ ( d − ) + patm ) D 2 ( H − ) (2)
4
4

patm = 101 kPa, γ = 9.80
kN
m
3
, H = 1 m, →
2
=
For d = 0, = 0, ➔ use negative sign:
− ( d + 11.31) + d (1m) = 0
( d + 11.31) ± d 2 + 18.61d + 128
2
d + 11.31) − d 2 + 18.61 d + 128
(
=
2
Tabulated data with the corresponding plot are shown below.
0.080
Water rise, l (m)
0.000
0.007
0.016
0.024
0.033
0.041
0.049
0.057
0.065
0.073
0.080
0.070
0.060
Water rise, ℓ (m)
Depth, d (m)
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
0.800
0.900
0.1000
0.050
0.040
0.030
0.020
0.010
0.000
0.000
0.200
0.400
0.600
Depth, d (m)
0.800
1.000
Problem 2.68
The basic elements of a hydraulic press are shown in the figure below. The plunger has an
area of 1 in.2 , and a force, F1 , can be applied to the plunger through a lever mechanism
having a mechanical advantage of 8 to 1. If the large piston has an area of 150 in.2 , what
load, F2, can be raised by a force of 30 lb applied to the lever? Neglect the hydrostatic
pressure variation.
F2
Plunger
Hydraulic fluid
F1
Solution 2.68
A force of 30 lb applied to the level results in a plunger force, F1 , of F1 = (8)(30) = 240 lb.
Since F1 = pA1 and F2 = pA2 where p is the pressure and A1 and A2 are the areas of the
plunger and piston, respectively. Since p is constant throughout the chamber,
F1 F2
=
A1 A2
 150 in.2 
A2
F1 = 
240 lb ) →
so that F2 =
 1in.2  (
A1


F2 = 36,000 lb
Problem 2.69
The hydraulic cylinder shown in the figure below, with a 4-in.- diameter piston, is
advertised as being capable of providing a force of F = 20 tons. If the piston has a design
pressure (the maximum pressure at which the cylinder should safely operate) of 2500 lb/in2 ,
gage, can the cylinder safely provide the advertised force?
patm
pmax = 2500 Ib/in.2
gage
F
Solution 2.69
Assuming a “ton” is a “short ton”, the advertised force is
 2000 lb 
Fadv = 20 tons 
 = 40000 lb
 ton 
The maximum force that can be safely developed by the piston is
lb  π
π
 
2
Fsafe = pgage A = pgage  d 2  =  2500 2  ( 4 in )  = 31400 lb
4  

in  4
No. The cylinder cannot safely provide 20 tons of force.
Problem 2.70
A Bourdon gage is often used to
Bourdon gage
𝒲
measure pressure. One way to
0.4-in.-diameter
calibrate this type of gage is to use
the arrangement shown in the figure
θ
Liquid
below (a). The container is filled
with a liquid and a weight, W ,
placed on one side with the gage on
(a)
(b)
the other side. The weight acting on
the liquid through a 0.4-in. -diameter opening creates a pressure that is transmitted to the
gage. This arrangement, with a series of weights, can be used to determine what a change in
the dial movement, θ , in the figure below (b), corresponds to in terms of a change in
pressure. For a particular gage, some data are given below. Based on a plot of these data,
determine the relationship between θ and the pressure, p, where p is measured inpsi.
0
1.04
2.00
3.23
4.05
5.24
6.31
(deg.)
0
20
40
60
80
100
120
Solution 2.70
p=
W ( lb )
W
=
= 7.96 W ( lb )
Area π 0.4 in. 2
(
)
4
(1)
Weight, lb
 (lb)
W = 0.0522 θ
So that from Eq. (1)
p ( psi )
= 0.0522 θ
7.96
p ( psi ) = 0.416 θ
W = 0.0522 θ
0
50
100
Theta, degrees
(where p is in psi )
From graph
8.00
6.00
4.00
2.00
0.00
Theta, deg
0
20
40
60
80
100
120
W, lb
0.00
1.04
2.00
3.23
4.05
5.24
6.31
150
Problem 2.71
A bottle jack allows an average person to lift one corner of a 4000-lb automobile
completely off the ground by exerting less than 20 lb of force. Explain how a 20-lb force
can be converted into hundreds or thousands of pounds of force, and why this does not
violate our general perception that you cannot get something for nothing (a somewhat
loose paraphrase of the first law of thermodynamics). Hint: Consider the work done by
each force.
Solution 2.71
Known: 20 lb applied force lifts corner of 4,000 lb automobile
Determine: Explain
Strategy: Force = (pressure)(area); consider work done by pistons of different size
Solution:
Consider two piston/cylinders of
different diameter connected by a rigid
tube filled with a liquid.
F1
F
= p= 2
A1
A2
A1
F1
A 
d 
→ F2 = F1  2  = F1  2 
 A1 
 d1 
d2
= 4 → F2 = 16F1
d1
p
A2
2
d2
= 10 → F2 = 100F1
d1
Therefore producing the required force multiplication is not difficult.
Assuming all solid boundaries are rigid, the volume pushed out of the small cylinder
must equal that entering the large cylinder.
Δ V 1 = Δx1A1 = Δx2 A2
A 
→ Δx2 = Δx1  1 
 A2 
Therefore, the larger piston moves a smaller distance than the smaller piston.
Comparing the work done on the smaller piston to the work done by the larger piston:
  A  
 A 
W2 = F2 Δx2 =  F1  2    Δx1  1   = F1Δx1 = W1
 A2  
  A1   
Therefore, the work done on the small piston is equal to work done by large piston.
F2
Problem 2.72
Suction is often used in manufacturing processes to lift objects to be moved to a new
1
location. A 4-ft by 8-ft sheet of -in. plywood weighs approximately 36 lb . If the
2
machine’s end effector has a diameter of 5 in. , determine the suction pressure required to
lift the sheet, expressed in inches of H2 O suction.
Solution 2.72
Known: W = 36 lb ; DCUP = 5 in.
Determine: Suction required to lift sheet in inches H20
Strategy: Force = (pressure)(area);
Solution:
W = F = pA = pπ R 2
W
36 lb
lb
p=
=
= 1.833 2 = γ H2O h
2
2
in
πR
π ( 2.5 in )
lb
in2
h=
= 50.77 in H2 O
lb
1 ft3
62.4 3 ×
ft 1728 in3
1.833
h = 51 in H2 O
Problem 2.73
A piston having a cross-sectional area of 0.07 m 2 is
located in a cylinder containing water as shown in the
figure below. An open U-tube manometer is connected
to the cylinder as shown. For h1 = 60 mm and
h = 100 mm, what is the value of the applied force, P,
acting on the piston? The weight of the piston is
negligible.
P
Piston
h
Water
h1
Mercury
Solution 2.73
For equilibrium, p = p p AP where p p is the pressure acting on piston and AP is the area of
the piston. Also,
p p + γ H2O h1 − γ Hg h = 0
kN 
kN 
kN


p p = γ Hg h − γ H2O h1 =  133 3  ( 0.100 m ) −  9.80 3  ( 0.060 m ) = 12.7 2
m 
m 
m


(
N 

p = 12.7 × 103 2  0.07 m 2
m 

) →
p = 889 N
Problem 2.74
A 6-in.-diameter piston is located within a cylinder that is connected to a
inclined-tube manometer as shown in the
figure below. The fluid in the cylinder and
the manometer is oil (specific weight =
59 lb/ft3 ). When a weight, W , is placed on
the top of the cylinder, the fluid level in the
manometer tube rises from point (1) to (2).
How heavy is the weight? Assume that the
change in position of the piston is negligible.
Solution 2.74
With piston alone let pressure on face of
piston = p p . Manometer equation becomes
p p − γ oil h1 sin30° = 0
𝒲
W
AP
Oil
(AP ~ area of piston)
(2)
Subtract Eq. (1) from Eq. (2) to obtain
6 
p′p − p p − γ oil  ft  sin30° = 0
 12 
W
6 
= γ oil  ft  sin30°
AP
 12 
 lb  6 
=  59 3   ft  ( 0.5 ) →
π  6   ft   12 
 ft 
4  12 
W
2
30°
𝒲
Piston
W = 2.90 lb
(2)
(1)
Oil
(1)
6 

p′p − γ oil  h1 + ft  sin30° = 0
12 

6 in.
Piston
With weight added pressure p p increases to p′p where
p′p = p p +
1
-in. -diameter
2
6 in.
h1
(2)
(1)
30°
Problem 2.75
The container shown in the figure below has square cross sections.
Find the vertical force on the horizontal surface, ABCD.
2
A
B
1
_″
22
C
D
1
_″
Water
42
E
F
4
Solution 2.75
R
The vertical force on surface ABCD is equal to the weight of the
imaginary fluid above ABCD as show on the picture on the right, so
F = γWV
r
A
(
)
lbm 

=  62.4 3  π ( 22 − 12 ) ft 2 ( 2.5 ft ) →
ft 

= γW π R 2 − r 2 h
F = 1470 lb
B
C
D
Problem 2.76
Find the weight W needed to hold the wall shown in the figure below upright. The wall is
10 m wide.
3m
𝒲
Wall
Water
4m
Pinned
Solution 2.76
The hydrostatic force F on the wall is found from
Fw
F = ρ ghc A
(
kg  
m

= 1000 3   9.81 2  ( 2m ) 4 × 10m 2

m 
s 
)
 kg ⋅ m   kN 
= 78500  2  

 s   1000 N 
= 785 kN
H
F
h
The force F is located one-third of the water depth from the bottom of the water.
h=
1
( 4m ) = 1.33 m
3
Summing moments about the pinned joint,
FW =
(1.33 m ) 785 kN = 149 kN
h
F=
(
)
H
(7 m )
Assuming no friction between the rope and the pulley,
W = FW
→
W = 149 kN
DISCUSSION
Note that the atmospheric pressure acts on both sides of the wall.
Therefore, the forces due to atmospheric pressure are equal and opposite, and cancel.
Problem 2.77
Determine the magnitude and direction of the force that must
be applied to the bottom of the gate shown in the figure below
to keep the gate closed.
Water
2.1 m
Hinge
2-m-wide
gate
0.8 m
Solution 2.77
The hydrostatic force on the gate is
F = γ yc A
kg  
m

= 1000 3   9.81 2  (1.3 m + 0.4 m )( 2 m × 0.8 m )
m 
s 

= 26700 N
H
yc y
p
yR
h
The location of the force F is
y p = yc +
F
I xc
12 yc A
FR
Using Appendix,
y p = yc +
bh3
h2
= yc +
12 yc A
12 yc
2
0.8 m )
(
= (1.3 + 0.4 ) m +
= 1.73 m
12 (1.3 + 0.4 ) m
Summing moments about the hinge,
 M hinge = FR h − F ( yp − H ) = 0
FR =
(
F yp − H
h
) = ( 26700N )(1.73 − 1.3) m →
0.8 m
FR = 14, 400 N
Problem 2.78
An automobile has just dropped into a river. The car door is approximately a rectangle,
measures 36 in. wide and 40 in. high, and hinges on a vertical side. The water level inside
the car is up to the midheight of the door, and the air inside the car is at atmospheric
pressure. Calculate the force required to open the door if the force is applied 24 in. from
the hinge line. See the figure below. (The driver did not have the presence of mind to open
the window to escape.)
4′
36′′
24′′
40′′
Solution 2.78
Note that the force due to atmospheric pressure acts in equal and opposite directions on
two sides of the door. The hydrostatic force on the inside of the door is
lb   ft3 

 36 in. × 40 in. 
10 in ) 
Fi = γ hc A =  62.4 3  
(

 = 260 lb
3
2


ft   1728 in 

Fo
The hydrostatic force on the outside of the door is
hinge
Fo = γ hc A
lb   ft3  
12 in


4ft
=  62.4 3  
×
+ 20in  ( 36 × 40 ) in2


3

ft
ft   1728 in  


= 3540 lb
18″
Fi
24″
Freq.
Top view of door
Summing moments about the hinge line
 M H = ( 24 in.) Freq + (18 in.) Fi − (18 in.) Fo
 18 
 18 
Freq = ( Fo − Fi )   = ( 3540 − 260 ) lb   →
 24 
 24 
Freq = 2, 460 lb
Problem 2.79
Consider the gate shown in the figure below. The gate is massless and has a width b
(perpendicular to the paper). The hydrostatic pressure on the vertical side creates a
counterclockwise moment about the hinge, and the hydrostatic pressure on the horizontal
side (or bottom) creates a clockwise moment about the hinge. Show that the net clockwise
moment is
 2 h2 
u
ghb
=
ρ
 w  2 − 6  .


h
ℓ
Hinge
Water, pw
Solution 2.79
The vertical force on the horizontal side is
Fv = ρw ghA = ρw gh ( l × b )
Constant force resultant acts at midpoint
The horizontal force on the vertical side is
h
FH = ρw ghc A = ρw g   ( h × b )
2
The resultant acts at
1 3
I xc h 12 bh
h h 2h
y p = yc +
= +
= + =
yc A 2 h bh
( ) 2 6 3
2
Summing moments about the hinge
→
 l  ρ gh b  h 
+  M =Fv x p − FH h − y p = ρw ghlb   − w
 
2
2
3
(
→
)
 l 2 h2 
+  M = ρw ghb  − 
2 6 


→
2
Problem 2.80
Consider the gate shown in the figure below. The gate is massless and has a width b
(perpendicular to the paper). The hydrostatic pressure on the vertical side creates a
counterclockwise moment about the hinge, and the hydrostatic pressure on the horizontal
side (or bottom) creates a clockwise moment about the hinge. Will the gate ever open?
h
ℓ
Hinge
Water, pw
Solution 2.80
Sum moments about hinge
→
+  M = FV LV − FH LH
l
h h
= ρw ghbl   − ρw g   bh  
2
2 3
 l 2 h2 
= ρw ghb  − 
2 6 


h2 l 2
> ,
6
2
or
h > 3l
If sum of moments is negative, gate will open
→
+ M < 0 →
h2 l 2
>
6
2
→
if h > 3 l , gate will open
Problem 2.81
A tank contains 6 in. of oil ( SG = 0.82 ) above 6 in. of water ( SG = 1.00 ). Find the force
on the bottom of the tank. See the figure below.
A = 1 ft2
Oil
(S = 0.82)
Water
(S = 1.0)
Solution 2.81
Assume atmospheric pressure acts on outside of tank.
Pressure is constant at constant elevation in a stagnant fluid.
Apply hydrostatic pressure equation
FNET = pBOTTOM A
= ( γ oil hoil + γ w hw ) A
= γ w ( SGoil hoil + SGw hw ) A
lb  

 1   1 
=  62.4 3  0.82  ft  + 1 ft   1 ft 2
ft  
 2   2 

( ) →
FNET = 56.7 lb
Problem 2.82
A structure is attached to the ocean floor as shown in the figure below. A 2-m -diameter
hatch is located in an inclined wall and hinged on one edge. Determine the minimum air
pressure, p1 , within the container that will open the hatch. Neglect the weight of the hatch
and friction in the hinge.
Free surface
Seawater
10 m
30°
Hatch
Hinge
Air pressure, p1
Solution 2.82
FR = γ hc A where hc = 10 m +
1
( 2 m ) sin30° = 10.5 m
2
Hy
Hx
FR
Thus,
N 

π 
2
FR = 10.1× 103 3  (10.5 m )   ( 2 m ) = 3.33 × 105 N
m 

4
To locate FR,
yR =
I xc
10 m
+ yc where yc =
+ 1m = 21m
yc A
sin30°
4
π 
  (1m )
4
yR =  
+ 21 m = 21.012 m
( 21m )(π )(1m )2
For equilibrium,
 MH = 0
FR ( 21.012 m − 20 m ) = p1 (π )(1m ) (1m )
2
3.33 × 105 N ) (1.012 m )
(
p =
→
1
π (1m ) (1m )
2
p1 = 107 kPa
p1 A
Problem 2.83
Concrete is poured into the forms as shown in the figure below to produce a set of steps.
Determine the weight of the sandbag needed to keep the bottomless forms from lifting off
the ground. The weight of the forms is 85 lb , and the specific weight of the concrete is
150 lb / ft3.
Open top
3 ft
8 in. risers
Sand
Open bottom
10 in. tread
Solution 2.83
From the free-body-diagram
𝒲s
( ↓ )  Fy = 0
+
𝒲c
Ws +Wc + W f − pb A = 0
𝒲f
(1)
pb A
Ws = weight of sandbag
Wc = weight of concrete
W f = weight of forms
pb = pressure along bottom surface due to concrete
A = area of bottom surface
From the data given:
lb 

Wc =  150 3  ( Vol. concrete )
ft 

(10 in.)( 24 in.) + (10 in.)(16 in.) + (10 in.)( 8 in.) 
lb 

=  150 3  ( 3ft ) 
= 1500 lb
ft 
in.2

144 2
ft
W f = 85 lb
lb   24 
lb

pA = 150 3   ft  = 300 2
ft   12 
ft

 30 
A =  ft  ( 3ft ) = 7.5 ft 2
 12 
Thus, from Eq. (1)
(
)
lb 

Ws =  300 2  7.5 ft 2 − 1500 lb − 85 lb →
ft 

Ws = 665 lb
y
Problem 2.84
A long, vertical wall separates seawater from fresh water. If the seawater stands at a depth
of 7 m, what depth of freshwater is required to give a zero resultant force on the wall?
When the resultant force is zero, will the moment due to the fluid forces be zero? Explain.
Solution 2.84
For a zero resultant force
FRs = FRf
Fresh
water
(γf )
Sea water
(γ s )
γ s hcs As = γ f hcf Af
h
7m
Thus, for a unit length of wall
FRf
FRs
kN   7 m 
kN   h 


10.1 3  
 ( 7 m × 1m ) =  9.80 3   m  ( h × 1m )
m  2 
m  2 


0
h = 7.11 m
In order for moment to be zero, FRs and FRf must collinear.
1
(1m )( 7 m )3 7
I xc
+ yc = 12
+ m = 4.67 m
For FRs : yR =
yc A
2
7 
 m  ( 7 m ×1m )
2 
1
3
1m
7.11m
(
)(
)
I
7.11
+
m = 4.74 m
Similarly for FRf : yR = xc + yc = 12
yc A
2
 7.11 
m  ( 7.11m × 1m )

 2

Thus, the distance to FRs from the bottom (point 0) is 7 m − 4.67 m = 2.33 m .
For FRf this distance is 7.11m − 4.74 m = 2.37 m .
The forces are not collinear.
→ No; for zero resultant force, the sum of the moments will not be zero.
Problem 2.85
Forms used to make a concrete basement wall are shown in the
figure below. Each 4-ft -long form is held together by four
ties—two at the top and two at the bottom as indicated.
Determine the tension in the upper and lower ties. Assume
concrete acts as a fluid with a weight of150 lb / ft3.
10 in.
1 ft
Tie
Concrete
10 ft
Form
1 ft
Solution 2.85
F1
(1)  Fx = 0, or F1 + F2 = FR
1 ft
hc = 5 ft
and
(2) M 0 = 0, or 1F1 + 2 F2 =
R FR , where FR = pc A = γ hc A
C
FR
ℓ1 = 9 ft
Thus,
FR = 150
ℓR
lb
ft
( 5 ft )(10 ft )( 4 ft ) = 30000 lb
3
F2
Z
ℓ2 =1 ft
X
Also,
0
Width = 4 ft
1
( 4 ft )(10 ft )3
I xc
= 10 ft − 5 ft − 12
R = 10 ft − yR = 10 ft − yc − ( yR − yc ) = 10 ft − hc −
yc A
5 ft (10 ft )( 4 ft )
= 5 ft − 1.67 ft = 3.33 ft
Thus, from Eq. (2):
( 9 ft ) F1 + (1ft ) F2 = ( 3.33ft )(30000 lb ) = 99,900 ft ⋅ lb
(3)
9F1 + F2 = 99900
From Eq. (1), F1 + F2 = 30,000 lb, or F2 = 30000 − F1
9F1 + 30000 − F1 = 99,900
F1 = 8,740 lb
F2 = 30000 lb - 8740 lb
F1 = 21,260 lb
Problem 2.86
While building a high, tapered concrete
wall, builders used the wooden forms
shown in the figure below. If concrete
has a specific gravity of about 2.5, find
the total force on each of the three side
sections (A, B, and C) of the wooden
forms (neglect any restraining force of
the two ends of the forms).
Freshly poured
concrete
8 ft
A
6 ft
Wooden
forms
B
B
C
5 ft
12 ft
C
8 ft
Hardened
concrete
Solution 2.86
The horizontal force FA
5′
FA = γ yc A
FB
6′
lb 

= 2.5  62.4 3  ( 6 ft )(12 × 8 ) ft 2
ft 

12′
→
x
FCV
FA
FA = 89900 lb
FCH
8′
The force FB is horizontal
y
FB = γ yc A
lb 

= 2.5  62.4 3  ( 3ft )( 6 × 8 ) ft 2
ft 

→
FB = 22500 lb
The horizontal force FCH
lb 

FCH = γ yc A = 2.5  62.4 3  ( 9ft )( 6 × 8 ) ft 2 = 67400 lb
ft 

The vertical force FCV is the weight of concrete “above” the slanted side (the dashed
volume)
lb  
1


FCV = γV = 2.5  62.4 3  ( 3 × 6 × 8 ) ft3 + ( 3 × 6 × 8 ) ft3  = 33700 lb
2
ft  


The total force FC is
FC = FCH 2 + FCV 2 =
( 67400 )2 + ( 33700 )2 lb
→
FC = 75400 lb
Problem 2.87
A homogeneous, 4-ft -wide, 8-ft -long rectangular gate weighing 800 lb is held in place by a
horizontal flexible cable as shown in the figure below. Water acts against the gate, which is
hinged at point A. Friction in the hinge is negligible. Determine the tension in the cable.
Cable
Water
60°
Gate
6 ft
8 ft
Hinge
A
Solution 2.87
 6 ft 
FR = γ hc A where hc = 
 sin 60°
 2 
T
Thus,
FR
lb   6 ft 

FR =  62.4 3  
 ( sin 60° )( 6 ft × 4 ft ) = 3890 lb
ft   2 

Hy
To locate FR ,
yR =
I xc
+ yc where yc = 3ft
yc A
so that
1
( 4 ft )( 6 ft )3
yR = 12
+ 3ft = 4.0 ft
( 3ft )( 6 ft × 4 ft )
For equilibrium,
 MH = 0
and
T ( 8ft )( sin 60° ) = W ( 4 ft )( cos 60° ) + FR ( 2 ft )
T=
( 800 lb )( 4 ft )( cos 60° ) + (3890 lb )( 2 ft )
( 8ft )( sin 60° )
→
T = 1350 lb
𝒲
Hx
Problem 2.88
A gate having the shape shown in the figure below is
located in the vertical side of an open tank
containing water. The gate is mounted on a
Water
horizontal shaft. (a) When the water level is at the
top of the gate, determine the magnitude of the fluid
force on the rectangular portion of the gate above
the shaft and the magnitude of the fluid force on
the semicircular portion of the gate below the shaft.
(b) For this same fluid depth determine the moment
of the force acting on the semicircular portion of the
gate with respect to an axis that coincides with the shaft.
6m
Shaft
3m
Side view
of gate
Solution 2.88
(a) For rectangular portion,
( FR )r = γ hc A where hc = 3 m
c
N
( FR )r =  9800 3  ( 3 m )( 6 m × 6 m )
m 

a
––
2
x
a
––
2
y
b
––
2
( FR )r = 1060 kN
12
1 ab3
Iyc = –––
Ixyc = 0
π R2
A = –––––
2
Ixc = 0.1098R4
( FR )sc = γ hc A where
c
y
4 (3 m )
4R
hc = 6 m +
= 6m +
= 7.27 m
3π
3π

1 ba3
Ixc = –––
12
b
––
2
For semi-circular portion,
( FR )sc =  9800
A = ba
R
x
R
4R
–––
3π
Iyc = 0.3927R4
Ixyc = 0
N 
π
2
7.27 m )  ( 3 m ) 
3 (
m 
2

( FR )sc = 1010 kN
(b) For semi-circular portion yR =
I xc
+ yc =
yc A
0.1098 R 4
+ 7.27 m=7.36 m
π  2
( 7.27 m )   R
2
Thus, moment with respect to shaft,M :
(
)
M = ( FR )sc × ( 7.36 m − 6.00 m ) = 1010 × 103 N (1.36 m ) →
M = 1.37 × 106 N ⋅ m
Problem 2.89
A pump supplies water under pressure to a large tank as shown in the figure below. The
circular-plate valve fitted in the short discharge pipe on the tank pivots about its diameter
A–A and is held shut against the water pressure by a latch at B. Show that the force on the
latch is independent of the supply pressure,p, and the height of the tank, h.
Pressure p
Supply
Water
h
A
A
B
Solution 2.89
The pressure on the gate is the same as it would be for an open tank with a depth of
hc =
p+γh
γ
as shown in the figure.
 M A = 0, or
(1) ( yR − yc ) FR = RFB
p/γ
Pressure p
hc
Water
h
A
A
B
2R
Supply
where
(
)
(
FR = pc A = γ hc π R 2 = ( p + γ h ) π R 2
)
yc = hc
and
C
π R4
(2) yR − yc =
I xc
R2
4
=
=
yc A  p + γ h 
p

2
4 + h
 γ π R


γ

yR – yc
( yR − yc )
FB = γ
R
π
4
FR =
(
R
( p + γ h ) π R2
p

4 + h
γ

FR
FB
Thus, from Eqs. (1) and (2)
FB =
A
)
R3 , which is independent of both p and h.
Problem 2.91
Find the center of pressure of an elliptical area of minor axis 2a and major axis 2b where
axis 2a is vertical and axis 2b is horizontal. The center of the ellipse is a vertical distance h
below the surface of the water ( h > a ). The fluid density is constant. Will the center of
pressure of the ellipse change if the fluid is replaced by another constant-density fluid? Will
the center of pressure of the ellipse change if the vertical axis is tilted back an angle α from
the vertical about its horizontal axis? Explain.
Solution 2.91
For a hydrostatic pressure distribution, using geometric information from the Appendix,
 π ba3 


I xc  4  a 2
=
=
y p − yc =
yc A h (π ab ) 4h
y p = yc +
x
y
h
a2
a2
= h+
4h
4h
2a
2b
Recognizing symmetry about minor axis,
xp = 0
Above expressions for x p and y p contain only geometric properties (and not fluid
properties)
Location of center of pressure not dependent on density.
Consider the side view of the ellipse.
x
The equation
y p − yc =
I xc
yc A
Requires that the y-coordinate lie in the plane of
the surface.
yc = h
yp – yc
α
Center of
pressure
I xc is smaller for the smaller horizontally projected area of the tilted ellipse.
Therefore, for the tilted ellipse,
( y p − yc ) is smaller for the tilted ellipse, the center of pressure is higher.
yp – yc
Problem 2.92
The dam shown in the figure below is 200 ft long and is made of concrete with a specific
gravity of 2.2. Find the magnitude and y coordinate of the line of action of the net
horizontal force.
20′
X
Headwater
g
y
60′
Tailwater
20′
40′
Solution 2.92
The headwater horizontal force and its line of action are
lb 

FHH = γ yc A =  62.4 3  ( 30 ft )( 60 × 200 ) ft 2 = 2.25 ×107 lb
ft 

yAP =
2
2
hH = ( 60 ft ) = 40 ft .
3
3
The tailwater horizontal force and its line of action are
lb 

FTH = γ yc A =  62.4 3  (10 ft )( 20 ft × 200 ft ) = 2.50 ×106 lb
ft 

yTP =
2
2
h T = ( 20 ft ) = 13.3 ft.
3
3
Fnet = FHH − FTH = 2.25 × 107 lb − 2.50 ×106 lb
FHH
Fnet = 2.00 ×107 lb acting to right.
The line if action is located by taking moments about the
base of the dam.
2.25 × 107 lb ) ( 20 ft ) − ( 2.50 × 106 lb ) ( 6.7 ft )
(
ynet =
( 2.00 ×107 lb )
ynet = 21.7 ft above base
60′–yHP
= 20′
20′–yTP
= 6.7′
Problem 2.93
The dam shown in the figure below is 200 ft long and is made of concrete with a specific
gravity of 2.2 . Find the magnitude and x coordinate of the line of action of the vertical
force on the dam resulting from the water.
20′
X
Headwater
g
y
60′
Tailwater
20′
40′
Solution 2.93
The only vertical force due to the water is on the headwater side of the dam. This vertical
force equals the weight of the water above the surface and the force acts through the
centroid of the water volume. Therefore
lb   1 

FHV = γ V =  62.4 3    ( 40 ft )( 60 ft )( 200 ft )
ft   2 

FHV = 1.50 ×107 lb
The x -location of this force is the centroid of the 40 ft × 60 ft triangle which gives
xp =
2
( 40 ft ) or x p = 26.7 ft
3
.
Problem 2.94
The figure below is a representation of the
Keswick gravity dam in California. Find the
magnitudes and locations of the hydrostatic
forces acting on the headwater vertical wall
of the dam and on the tailwater inclined wall
of the dam. Note that the slope given is the
ratio of the run to the rise. Consider a unit
length of the dam ( b = 1ft ).
21.5′
Headwater
El. 595.00′
El. 565.40′
100′
Slope – 0.7:1
Tailwater
30′
El. 491.00′
Solution 2.94
Consider a unit length of the dam. The headwater force is
γ bhH
h 
FHH = γ yc A = γ  H  ( bhH ) =
2
 2 
2
lb 
2

 62.4 3  (1ft )(100 ft )
ft 
=
2
FHH = 312,000 lb
The location y p is
yp =
2
2
hH = (100 ft )
3
3
yp
hH
y'p
y p = 66.7 ft
FHH
FT hT
The tailwater force is
 hT 
FT = γ yc A = γ 
A
 2 
where
A = ( 30 ft )(1ft )
1.02 + 0.72
= 36.6 ft 2
1.0
so
(
lb 

2
 62.4 3  ( 30 ft ) 36.6 ft
ft 
FT = 
2
)
or
FT = 34,300 lb
The location y′p is
y′p =
2
1.02 + 0.72 2
1.02 + 0.72
hT
= ( 30 ft )
3
1.0
3
1.0
or
y′p = 24.4 ft
.
Problem 2.95
The Keswick dam shown in the figure below is made of concrete and has a specific weight
of 150 lb/ft3. The hydrostatic forces and the weight of the dam produce a total vertical force
of the dam on the foundation. Find the magnitude and location of this total vertical force.
Consider a unit length of the dam ( b = 1 ft).
Headwater
21.5′
El. 595.00′
El. 565.40′
100′
Slope – 0.7:1
Tailwater
El. 491.00′
30′
Solution 2.95
The hydrostatic vertical force is due to the tailwater. Its magnitude
for a dam unit length is
1
FTV = γ V = γ   ( b ) h T ( 0.7h T ) = 0.35γ bhT 2
2
lb 

2
= 0.35  62.4 3  (1ft )( 30 ft )
ft 

FTV = 19700 lb
FTV
W1
W2
w1
The dam weight consists of W1 + W2 . Now
lb 

W1 = γ C V =  150 3  (1ft )( 21.5 ft )( 595.0 − 491.0 ) ft = 335400 lb
ft 

lb 

1
 52.1 
W2 = γ c V2 = 150 3  (1ft )   ( 52.1ft ) 
ft  = 290800 lb.
ft 
2
 0.7 

The total force F is
F = FTV +W1 + W2 = (19700 + 335400 + 290800 ) lb
F = 645,900 lb
.
w2
w1 = 21.5 ft.
w2 = 0.7(565.4 – 491.0) ft
= 52.1 ft.
Problem 2.96
The Keswick dam shown in the figure
below is made of concrete and has a specific
Headwater
21.5′
El. 595.00′
weight of 150 lb/ft3 . The coefficient of
friction μ between the base of the dam and
the foundation is 0.65 . Is the dam likely to
slide downstream? Consider a unit length of
the dam ( b = 1 ft ).
El. 565.40′
100′
Slope – 0.7:1
Tailwater
30′
El. 491.00′
Solution 2.96
The total vertical force acting downward is
F = W1 + W2 + FTV .
Using the result of Problem 2.51,
F = 645900 lb.
FHH
FTV
W1
W2
The horizontal force resisting movement of the dam is
μ F = 0.65 ( 645900 lb ) = 419800 lb
The net force causing the dam to move downstream is
( FHH − FTH ) . Using the result of Problem 2.50,


1.0
1.0
= 28100 lb
FTH = FT 
 = ( 34300 lb )

2
2 
2
2
1.0 + 0.7
 1.0 + 0.7 
Then
FHH − FTH = ( 312000 − 28100 ) lb = 283900 lb.
Since
FHH − FTH < μ F ,
the dam will not
slide downstream.
μF
FT
FTH
Problem 2.97
The figure below is a representation of the Altus
gravity dam in Oklahoma. Find the magnitudes
and locations of the horizontal and vertical
hydrostatic force components acting on the
headwater wall of the dam and on the tailwater
wall of the dam. Note that the slope given is the
ratio of the run to the rise. Consider a unit length
of the dam ( b = 1ft).
10′
EI. 1564.00′
EI. 1555.00′
EI. 1553.00′
Headwater
87′ Slope =
0.1:1
Slope = 0.6:1
EI. 1475.00
Tailwater
4′
Solution 2.97
First consider the headwater hydrostatic force components.
h 
FHH = γ yc A = γ  H  ( bhH )
 2 
=
x
2
γ bhH
yp
lb 

2
 62.4 3  (1ft )( 87 ft )
ft 
=
2
FHH = 236,200 lb
yp =
y
2
FTV
w1 = 0.1(1555.0 − 1475.0 ) ft
=8ft.
.
FHV = γ V ' + γ V "
lb 

 (1ft )( 7 ft )( 87 ft )
ft3 

= 3490 lb
γ V ' =  62.4
FHV
w1
.
2
2
hH = ( 87 ft )
3
3
y p = 58 ft
FHH
7′
V'
V"
FTH
lb  (1ft )( 8ft )( 87 − 7 ) ft


2
ft3 

= 19970 lb
γ V " =  62.4
FHV = 3490 lb + 19970 lb
FHV = 23460 lb
.
The location x p of FHV is found from x p =
γ V ' x'p + γ V '' x''p
γ V ' + γ V ''
where x'p and x''p are the horizontal locations of γ V ' and γ V '' respectively.
The numerical values give
xp =
( 3490 lb )  × 8ft  + (19970 lb )  × 8ft 
1
2

3490
+ 19970 ) lb
(
x p = 5.14 ft
2
3

.
The tailwater hydrostatic force components
xp
h 
FTH = γ yc A = γ  T  ( bhT )
 2 
=
FTV
γ bhT2
yp
hT
2
lb 

2
 62.4 3  (1ft )( 4 ft )
ft 
=
2
FTH
→
FTH = 499 lb
ℓ
= 0.6 ( 4 ft ) = 2.4 ft.
yp =
2
2
hT = ( 4 ft ) or
3
3
y p = 2.67 ft
.
lb 

62.4 3  (1ft )( 2.4 ft )( 4 ft )

 h  γ b hT 
ft 
=
FTV = γ V = γ b  T  =
2
2
 2 
xp =
2
2
= ( 2.4 ft ) or
3
3
x p = 1.6 ft
.
→
FTV = 300 lb
Problem 2.98
The Altus dam in the figure below is made of
concrete with a density of 150 lbm/ft3 . The
coefficient of friction μ between the base of the
dam and the foundation is 0.65. Is the dam likely
to slide downstream? Consider a unit length of
the dam ( b = 1 ft ).
10′
EI. 1564.00′
EI. 1555.00′
EI. 1553.00′
Headwater
87′ Slope =
0.1:1
Slope = 0.6:1
Tailwater
4′
EI. 1475.00
Solution 2.98
The total vertical force acting downward is
F = FHV + FTV +W1 + W2 +W3 .
Using the results of Problem 2.97
FH = 23, 460 lb
FTV = 300 lb .
The dam weights are
w1 = 0.1(1555.0 − 1475.0 ) ft=8ft
FHV
W1
W2
W3
w1
w3
w3 = 0.6 (1553.0 − 1475.0 ) ft = 46.8ft
lb   1 

W1 = γ c V 1 =  150 3    (1ft )( 8ft )(1555.0 − 1475.0 ) ft=48,000 lb
ft   2 

lb 

W2 = γ c V 2 =  150 3  (1ft )(10 ft )(1564.0 − 1475.0 ) ft = 133,500 lb
ft 

lb   1 

W3 = γ c V 3 = 150 3    (1ft )( 46.8ft )(1553.0 − 1475.0 ) ft = 273,800 lb.
ft   2 

F = ( 23460 + 300 + 48000 + 133500 + 273800 ) lb = 479,000 lb
FTV
Horizontal force resisting sliding movement of the dam is:
μ F = 0.65 ( 479000 lb ) = 311, 400 lb.
Net force acting to slide dam downstream is ( FHH − FTH ) . Using the results of Problem 2.53
( FHH − FTH ) = ( 236200 − 499 ) lb=235700 lb.
( FHH − FTH ) < μ F ,
the dam will not
slide downstream.
Problem 2.99
Find the magnitude and location of
the net horizontal force on the gate
shown in the figure below. The gate
width is 5.0 m.
1.0 m
A
4.0 m
Water
3.0 m
Water
2.0 m
45°
Solution 2.99
B
2.0 m
The hydrostatic horizontal force FH' is
FH' = γ w hc A = ρw ghc A
kg  
m

=  1000 3   9.81 2  ( 2.0 m )( 2.0 × 5.0 ) m 2

m 
s 
= 196000 N=196 kN
The location h ' of FH' is
h ' = 3.0 m − y p

I 
= 3.0 m −  yc + xc 
yc A 

1

( 5.0 m )( 2.0 m )3 

= 3.0 m −  2.0 m + 12

( 2.0 m )( 2.0 × 5.0 ) m 2 



= 0.833 m
The hydrostatic horizontal force FH'' is
FH'' = γ w hc A = ρw ghc A
kg  
m

=  1000 3   9.81 2  ( 3.0 m )( 2.0 × 5.0 ) m 2
m 
s 

= 294000 N = 294 kN
2.0 m
1.0 m
4.0 m
FH"
FH'
h"
h'
The location of h of FH is

I 
h '' = 4.0 m − y p = 4.0 m −  yc + xc 
yc A 

1

( 5.0 m )( 2.0 m )3 

= 4.0 m −  3.0 m + 12

( 3.0 m )( 2.0 × 5.0 ) m 2 



= 0.889 m.
The magnitude of the net horizontal force FH
FH = FH" − FH' = 294 kN -196 kN
FH = 98 kN
The location of the net horizontal force F above the base is denoted by h and is found by
noting the moment of the resultant is equal to the moment of the components or
Fh = FH'' h'' − FH' h' ,
FH'' h'' − FH' h'
F
( 294 kN )( 0.889 m ) − (196 kN )( 0.833 m )
=
98kN
h=
h = 1.00 m
.
DISCUSSION Note that the resultant of the two opposing forces is not located between
the two forces; this is only true for two forces that are acting in the same direction. Again
note that the atmospheric pressure force was not considered as it acts uniformly on both
sides of the gate and cancels out. Also note that the horizontal hydrostatic forces do not
depend on the 45o angle.
Problem 2.100
Find the magnitude and location of the net vertical force on the gate shown in the figure
below. The gate width is 5.0 m.
1.0 m
A
4.0 m
Water
Water
3.0 m
2.0 m
45°
B
2.0 m
Solution 2.100
B
The hydrostatic vertical force FV' is the weight of the
water above the gate to the level A-A.
2.0 m
Then
B
A
ℓ1
W1
ℓ2
W2
ℓ"
4.0 m
ℓ3
W3
A
F"V
F'V
ℓ'
ℓ = 2.0 m
FV' = W3 + W2 = ρw g ( V 3 + V 2 )
2
kg  
m  1

 N ⋅s
= 1000 3  9.81 2   ( 2.0 × 2.0 × 5.0 ) m3 + ( 2.0 ×1.0 × 5.0 ) m3 
m 
s  2

 kg ⋅ m
FV' = 98100 N+98100 N = 196200 N
1.0 m
3.0 m
The hydrostatic vertical force FV'' is the weight of the water above the gate to the level B-B.
FV'' = W3 +W2 + W1 = ρw g ( V 3 + V 2 + V 1 )
 N ⋅ s2 
kg  
m

FV'' = 98100 N+98100 N+ 1000 3   9.81 2  ( 2.0 × 1.0 × 5.0 ) m3 
 kg ⋅ m 

m 
s 


= 98100 N+98100 N+98100 N
=294300 N
The magnitude of the vertical force FV
FV = FV'' − FV' = 294300 − 196200 N
FV = 98100 N acting downward.
The location of F is found by first finding the locations of FV' and FV'' . First
'
=
W3 3 + W2 2
.
W3 +W2
Recognizing that
3 =
3
and
2 =
2
gives
 2.0 m 
 2.0 m 
98100 N 
 + 98100 N 

3 
2 


'
=
98100 N + 98100 N
'
= 0.833 m
Also
''
=
W3 3 + W2 2 + W1 1
W3 + W2 + W1
Recognizing that
1=
2
gives
 2.0 m 
 2.0 m 
 2.0 m 
98100 N 
 + 98100 N 
 + 98100 N 

3 
2 
2 



''
=
98100 N + 98100 N + 98100 N
''
= 0.889 m .
The location of the resultant force from the left side of the gate is denoted by
V and is
found from
FV'' '' − FV' '
V =
FV
=
294300 N ( 0.889 m ) − 196200 N ( 0.833 m )
294300 N − 196200 N
V = 1.00 m
from left side of gate.
DISCUSSION Noting that the resultant vertical force, FV = 98100 N is the weight of a
volume of water measuring = 2.0 m, width 5.0 m , and height = 1.0 m , is there a quick way
to find FV ?
Problem 2.101
Find the total vertical force on the cylinder shown in the figure below.
patm
D = 6 cm
18 cm
3 cm
5 cm
d=
3 cm
Water
Solution 2.101
The net force F on the cylinder is due to the water and is
F = F1 + F2 = p1A1 + p2 A2 .
Since the atmospheric pressure does not contribute to the net force,
p1 and p2 will be considered gage pressures.
p1 A1
kg  
m

1000 3  9.81 2  (18 − 5 ) cm
N
m 
s 
p1 = ρw gh = 
= 1275 2
cm   kg ⋅ m 

m
100

2 
m

 N ⋅s 
kg  
m

1000 3   9.81 2  ( 3 ) cm
N
m 
s 
p2 = ρw gh = 
= 294 2
cm   kg ⋅ m 

m
 100

2 
m

 N ⋅s 
Then
2
N π
m  
N π 2 2
m 
2

2
F = 1275 2  ( 3cm ) 
 +  294 2  6 − 3 cm 


m 4
m 4
 100 cm  
 100 cm 
or
F = 1.52 N .
(
)
2
p2 A2
Problem 2.102
A 3-m -wide, 8-m -high rectangular gate is located at the end of a rectangular passage that
is connected to a large open tank filled with water as shown in the figure below. The gate is
hinged at its bottom and held closed by a horizontal force, FH , located at the center of the
gate. The maximum value for FH is 3500 kN. (a) Determine the maximum water depth,
h, above the center of the gate that can exist without the gate opening. (b) Is the answer the
same if the gate is hinged at the top? Explain your answer.
h
4m
FH
4m
Hinge
Solution 2.102
For gate hinged at bottom
 MH = 0
yR h = y c
so that
FR
( 4 m ) FH = FH
(see figure) (1)
ℓ
Hx
and
Hy
kN 

FR = γ hc A =  9.80 3  ( h )( 3 m × 8 m ) = ( 9.80 × 24 h ) kN
m 

1
(3 m )( 8 m )3
I xc
5.33
yR =
+ yc = 12
+h=
+h
yc A
h (3 m × 8 m )
h
Thus,
FH
( m ) = h + 4 − 
5.33
5.33

+ h = 4 −
h
 h

and from Eq. (1)
( 4 m )( 3500 kN ) =  4 −

so that
h = 16.2 m
5.33 
 ( 9.80 × 24 )( h ) kN
h 
ℓ = h + 4 – yR
For gate hinged at top
 MH = 0
Hy
Hx
ℓ1
so that
( 4 m ) FH = 1FH
(see figure) (1)
FH
FR
ℓ1 = yR – (h – 4)
where
1 = yR −
( h − 4 ) = 
5.33
5.33

+ 4  − (h − 4) =
+4
h
 h

Thus, from Eq. (1)
( 4 m )( 3500 kN ) = 
5.33

+ 4  ( 9.80 × 24 )( h ) kN
 h

and
h = 13.5 m
Maximum depth for gate hinged at top is less than maximum depth for gate hinged at
bottom.
Problem 2.103
A gate having the cross section shown in the figure below is 4 ft wide and is hinged at C.
The gate weighs 18,000 lb, and its mass center is 1.67 ft to the right of the plane BC.
Determine the vertical reaction at A on the gate when the water level is 3 ft above the base.
All contact surfaces are smooth.
5 ft
Hinge
C
9 ft
Water surface
3 ft
B
A
Solution 2.103
F1 = γ hc A where hc = 1.5 ft
Cy
Cx
lb 

Thus, F1 =  62.4 3  (1.5 ft )( 3ft × 4 ft ) = 1120 lb
ft 

The force F1 acts at a distance of 1ft from the base of the gate.
F2 = p2 A2 where p2 = γ H2O ( 3ft )
Thus,
lb 

F2 =  62.4 3  ( 3ft )( 5 ft × 4 ft ) = 3740 lb
ft 

and acts at the center of the bottom gate surface.
For equilibrium,
 Mc = 0
and
F1 (11ft ) + F2 ( 2.5 ft ) + FA ( 5 ft ) = W (1.67 ft )
so that
FA =
(18,000 lb )(1.67 ft ) − (1120 lb )(11ft ) − ( 3740 lb )( 2.5 ft ) = 1680 lb
5 ft
F1
𝒲
F2
FA
Problem 2.104
The massless, 4-ft -wide gate shown in the figure below pivots about the frictionless hinge O.
It is held in place by the 2000 lb counterweight, W . Determine the water depth, h.
Water
Gate
h
Pivot O
Width = 4 ft
2 ft
3 ft
𝒲
Solution 2.104
FR = γ hc A where hc =
h
2
Thus,
FR
d
h
h2
FR = γ H2O ( h × b ) = γ H2O
( 4 ft )
2
2
Oy
Ox
To locate FR ,
( )
1
( 4 ft ) h3 h 2
I xc
12
yR =
+ yc =
+ = h
h
yc A
( 4 ft × h ) 2 3
2
b = 4 ft
For equilibrium,  M 0 = 0
FR d = W ( 3ft ) where d = h − yR =
h
=
3
so that
( 2000 lb )( 3ft )
2
Thus,
h3 =

2


(γ H O )  h2  ( 4 ft )
(3)( 2000 lb )( 3ft )
lb   1 

 62.4 3   2  ( 4 ft )
ft   

h = 5.24 ft
h
3
𝒲
Problem 2.105
A 200 − lb homogeneous gate 10 ft wide and 5 ft long is hinged at point A and held in place
by a 12-ft -long brace as shown in the figure below. As the bottom of the brace is moved to
the right, the water level remains at the top of the gate. The line of action of the force that
the brace exerts on the gate is along the brace. (a) Plot the magnitude of the force exerted
on the gate by the brace as a function of the angle of the gate, θ , for 0 ≤ θ ≤ 90° . (b) Repeat
the calculations for the case in which the weight of the gate is negligible. Comment on the
result as θ → 0 .
Moveable
stop
12 ft
5 ft
Gate
Water
A
θ
Brace
Solution 2.105
12 ft = L
ℓ = 5 ft
Gate
Water
A
θ
ϕ
FR
Moveable
stop
Brace
ϕ
ℓ
3
Hy
A
(a) For the free-body-diagram of the gate (see figure),
 FA = 0
so that
 


FR   +W  cos θ  = ( FB cos φ )( sin θ ) + ( FB sin φ )( cos θ ) (1)
3
2

Also,
sin θ = L sin φ
(assuming hinge and end of brace at same elevation)
or
sin φ =
L
sin θ
and
 sin θ 
FR = γ hc A = γ 
( w)
 2 
θ
Hx
𝒲
FB
where w is the gate width. Thus, Eq. (1) can be written as
 3
W
cos θ = FB ( cos φ sin θ + sin φ cos θ )
 ( sin θ ) w +
2
6
γ 
so that
 γ 2w 
 γ 2w 
W
W
θ
θ
sin
cos
+



 tan θ +
6 
2
6 
2
FB = 
=
cos φ sin θ + sin φ cos θ
cos φ tan θ + sin φ
For γ = 62.4
lb
ft3
,
(2)
= 5 ft, w = 10 ft, and W = 200 lb,
lb 

2
 62.4 3  ( 5 ft ) (10 ft )
200 lb
ft 

tan θ +
6
2 = 2600 tan θ + 100 (3)
FB =
cos φ tan θ + sin φ
cos φ tan θ + sin φ
Since sin φ =
sin φ =
L
sin θ and
= 5 ft , L = 12 ft
5
sin θ
12
and for a given θ , φ can be determined. Thus, Eq. (3)
can be used to determine FB for a given θ .
(b) For W = 0 , Eq. (3) reduces to
FB =
2600 tan θ
cos φ tan θ + sin φ
(4)
and Eq. (4) can be used to determine FB for a given θ . Tabulated data of FB vs. θ for both
W = 200 lb and W = 0 lb are given below.
θ (degree)
90.0
85.0
80.0
75.0
70.0
65.0
60.0
55.0
50.0
45.0
40.0
35.0
30.0
25.0
20.0
15.0
10.0
5.0
2.0
F(B) (lb) (W=200 lb)
2843
2745
2651
2563
2480
2403
2332
2269
2213
2165
2125
2094
2075
2069
2083
2130
2250
2646
3858
F(B) (lb) (W=0 lb)
2843
2736
2633
2536
2445
2360
2282
2210
2144
2085
2032
1985
1945
1911
1884
1863
1847
1838
1836
4500
4000
3500
Force, Ib
3000
W = 200 Ib
2500
2000
1500
W = 0 Ib
1000
500
0
0.0
10.0 20.0 30.0 40.0 50.0 60.0 70.0 80.0 90.0
Theta, deg
As θ → 0 the value of FB can be determined from Eq. (4),
FB =
2600 tan θ
cos φ tan θ + sin φ
Since
sin φ =
5
sin θ
12
it follows that
2
5 
cos φ = 1 − sin φ = 1 −   sin2 θ
 12 
2
and therefore
FB =
2600 tan θ
2
5
5 
1 −   sin2 θ tan θ + sin θ
12
 12 
=
2600
2
5
5 
1 −   sin2 θ + cos θ
12
 12 
Thus, as θ → 0
FB →
2600
= 1840 lb
5
1+
12
Physically this result means that for θ ≡ 0 , the value of FB is indeterminate, but for any
“very small” value of θ , FB will approach 1840 lb .
Problem 2.106
An open tank has a vertical partition and on one side contains gasoline with a density
ρ = 700 kg / m3 at a depth of 4 m , as shown in the figure below. A rectangular gate that is
4 m high and 2 m wide and hinged at one end is located in the partition. Water is slowly
added to the empty side of the tank. At what depth, h, will the gate start to open?
Partition
Stop
Water
4m
h
Gasoline
Hinge
Solution 2.106
FRg = γ g hcg Ag ; where g refers to gasoline.
kg  
m

FRg =  700 3   9.81 2  ( 2 m )( 4 m × 2 m )
m 
s 

FRg
FRw
= 110 ×103 N=110 kN
ℓw
FRw = γ w hcw Aw ; where w refers to water.
Hx
N  h 

FRw =  9.80 × 103 3    ( 2 m × h ) ; where h is depth of water.

m  2 
(
)
FRw = 9.80 ×103 h2
For equilibrium,  M H = 0 ➔ FRw w = FRg g
w =
h
;
3
ℓg
g =
4
m
3
(9.80 ×103 )( h2 )  3h  = (110 ×103 N )  34 m 
h = 3.55 m
which is the limiting value for h.
Hy
Problem 2.107
A horizontal 2-m-diameter conduit is half filled with a liquid ( SG = 1.6 ) and is capped at
both ends with plane vertical surfaces. The air pressure in the conduit above the liquid
surface is 200 kPa. Determine the resultant force of the fluid acting on one of the end caps,
and locate this force relative to the bottom of the conduit.
Solution 2.107
A1 ~ area of end
Fair
yR
Fresultant
d
Fliquid
0
Fair = pA,
A2 ~ area
covered by liquid
where p is air pressure
Thus,
N  π 

2
Fair =  200 × 103 2   ( 2 m ) = 200π ×103 N

m  4 
Fliquid = γ hc A2
where hc =
4R
3π
(see the figure below)
Thus,
N   4 (1m )   1   π 
2

3
Fliquid = (1.6 )  9.81× 103 3  
     ( 2 m ) = 10.5 ×10 N
3
2
4
π

m 
  
For Fliquid ,
yR =
I xc
+ yc
yc A2
where I xc = 0.1098R 4
and yc = hc =
4R
3π
(see the figure below)
πR
A = –––––
2
2
Ixc = 0.1098R4
c
y
R
x
4R
–––
3π
R
Iyc = 0.3927R4
Ixyc = 0
Thus,
yR =
0.1098 (1m )
4
 4 (1m )   1   π 
2

     (2 m)
 3π   2   4 
+
4 (1m )
= 0.5891m
3π
Since Fresultant = Fair + Fliquid = ( 200π + 10.5 ) ×103 N = 639 kN,
we can sum moments about O to locate resultant to obtain
Fresultant ( d ) = Fair (1m ) + Fliquid (1m − 0.5891m )
So that
200π ×103 N ) (1m ) + (10.5 × 103 N ) ( 0.4109 m )
(
d=
639 ×103 N
= 0.990 m above bottom of conduit.
Problem 2.108
A 4-ft by 3-ft massless rectangular gate is used to close the end of the water tank shown in
the figure below. A 200-lb weight attached to the arm of the gate at a distance ℓ from the
frictionless hinge is just sufficient to keep the gate closed when the water depth is 2 ft , that
is, when the water fills the semicircular lower portion of the tank. If the water were deeper,
the gate would open. Determine the distance .
ℓ
Hinge
Hinge
1 ft
200 lb
Gate
Water
2 ft
radius
3 ft
4 ft
Solution 2.108
Hy
ℓ
Hx
𝒲
ℓ1
yR
FR
ℓ1 = 1 ft + yR
FR = γ hc A
where hc =
4R
3π
(see the figure below)
Thus,
2
 4R   π R 
FR = γ H2O 



 3π   2 
2
lb   4 ( 2 ft )   π ( 2 ft ) 


 = 333 lb
=  62.4 3  


2
ft   3π  


To locate FR,
yR =
=
=
I xc
+ yc
yc A
0.1098 R 4
4R
+
2
 4R   π R  3π


 
 3π   2 
(see the figure below)
( 0.1098)( 2 ft )4 + 4 ( 2 ft ) = 1.178ft
2
3π
 4 ( 2 ft )  π ( 2 ft )

 3π


2
π R2
A = –––––
2
Ixc = 0.1098R4
c
y
R
x
4R
–––
3π
R
Iyc = 0.3927R4
Ixyc = 0
For equilibrium,
 MH = 0
W = FR (1ft+y R )
So that
And
=
(333 lb )(1ft + 1.178ft ) = 3.63ft
200 lb
Problem 2.109
A thin 4-ft-wide, right-angle gate with negligible mass is free to pivot about a frictionless
hinge at point O, as shown in the figure below. The horizontal portion of the gate covers a
1-ft-diameter drain pipe that contains air at atmospheric pressure. Determine the minimum
water depth, h, at which the gate will pivot to allow water to flow into the pipe.
Width = 4 ft
Right-angle gate
Water
h
Hinge
O
1-ft-diameter pipe
3 ft
Solution 2.109
FR
1
FR
2
ℓ1
Ox
Oy
ℓ2
For equilibrium,
 M0 = 0
FR1 × 1 = FR2 × 2
(1)
FR1 = γ hc1 A1
lb   h 

=  62.4 3    ( 4 ft × h ) = 125 h2
ft   2 

For the force on the horizontal portion of the gate (which is balanced by pressure on both
sides except for the area of the pipe)
lb 
2 
2
π 
π 
FR2 = γ h   (1ft ) =  62.4 3  ( h )   (1ft )
ft 
4
4

= 49.0 h
Thus, from Eq. (1) with
1=
(125 h2 )  3h  = ( 49.0 h )(3ft )
h = 1.88ft
h
and
3
2 = 3ft
Problem 2.110
The closed vessel of the figure below contains water with an air pressure of 10 psi at the
water surface. One side of the vessel contains a spout that is closed by a 6-in.-diameter
circular gate that is hinged along one side as illustrated. The horizontal axis of the hinge is
located 10 ft below the water surface. Determine the minimum torque that must be applied
at the hinge to hold the gate shut. Neglect the weight of the gate and friction at the hinge.
10 psi
Air
10 ft
Axis
6-in.-diameter
gate
Water
3
4
Solution 2.110
10 ft
Hy
Hx
O
3
_
5
C
6
__
ft
12
F1
F2
3
4
Let F1 ~ force due to air pressure, and F2 ~ force due to hydrostatic pressure distribution of
water.
Thus,
lb  
in.2   π   6 

F1 = pair A = 10 2  144 2     ft  = 283 lb
ft   4   12 
 in.  
2
and
F2 = γ hc A
1  3   6  
where hc = 10 ft +     ft  = 10.15 ft
2  5   12  
So that
2
lb 

 π  6 
F2 =  62.4 3  (10.15 ft )    ft  = 124 lb
 4   12 
ft 

Also,
yR2 =
I xc
+ yc
yc A
where yc =
10 ft 1  6 
+  ft  = 16.92 ft
3
2  12 
5
So that
4
 π  3 
   ft 
 4   12 
yR2 =
+ 16.92 ft = 16.92 ft
2
 π  6 
(16.92 ft )    ft 
 4   12 
For equilibrium,
 M0 = 0
And



10 ft 
 3 
C = F1  ft  + F2  yR2 −
3 
 12 


5 




10 ft 
 3 
C = ( 283 lb )  ft  + (124 lb )  16.92 ft −
= 102 ft ⋅ lb
3 
 12 


5 

Problem 2.111
(a) Determine the horizontal hydrostatic force on the 2309-m-long Three Gorges Dam
when the average depth of the water against it is175 m . (b) If all of the 6.4 billion people on
Earth were to push horizontally against the Three Gorges Dam, could they generate
enough force to hold it in place? Support your answer with appropriate calculations.
Solution 2.111
(a)
N  175 m 

FR = γ hc A =  9.80 ×103 3 
 (175 m × 2309 m )

m  2 
= 3.46 × 1011 N
(b)
Required average force per person =
3.46 × 1011 N
6.4 ×109
N 
lb 
= 54.1
12.2

person 
person 
Yes. It is likely that enough force could be generated since required average force per
person is relatively small.
Problem 2.113
A 2-ft-diameter hemispherical plexiglass “bubble” is to be used as a special window on the
side of an above-ground swimming pool. The window is to be bolted onto the vertical wall
of the pool and faces outward, covering a 2-ft-diameter opening in the wall. The center of
the opening is 4 ft below the surface. Determine the horizontal and vertical components of
the force of the water on the hemisphere.
Solution 2.113
 Fx = 0 , or FH = FR = pc A
hc = 4 ft
FV
Thus,
FH = γ hc A = 62.4
lb
ft
3
2 ft
π
FH
( 4 ft ) ( 2 ft )2 = 784 lb ( to right )
4
and
 Fy = 0, or FV = W = γV = γ
4 π R3
,
3 2
FBD of water
where R = 1ft
FR
Thus,
lb  4π (1ft ) 
 = 131lb ( down on bubble )
FV = 62.4 3 

6
ft 

W
3
y
x
FV
FH
Problem 2.114
Consider the curved surface shown in the figure below (a) and (b). The two curved surfaces
are identical. How are the vertical forces on the two surfaces alike? How are they different?
Oil
Oil
R
(a)
R
(b)
Solution 2.114
In both cases, the magnitude of the vertical force is the weight of shaded section shown on
the right. In addition, the location of the vertical force is the same (the centroid of the
shaded section.) Therefore:
Alike: magnitude and
location of vertical
forces same.
R
However, the two vertical forces are different in that the force in (a) is
acting upward and the force in (b) is acting downward. Therefore:
Different: direction of
vertical forces
opposite.
Problem 2.115
The figure below shows a cross section of a submerged tunnel used by automobiles to travel
under a river. Find the magnitude and location of the resultant hydrostatic force on the
circular roof of the tunnel. The tunnel is 4 mi long.
g
50′
R = 20′
Solution 2.115
Due to symmetry, there is no net horizontal force on the roof. The vertical force is equal to
the weight of fluid above the tunnel. This vertical force acts through the centroid of the
fluid volume. Then for a tunnel length ,
π


F = γ V = γ  2Rh − R 2 
2


π
lb 
ft  

2

=  62.4 3  ( 4 mi )  5280
 2 ( 20 ft )( 70 ft ) − ( 20 ft ) 
mi  
2
ft 



F = 2.86 × 109 lb .
F
O
acts downward
This force
through the
point "O".
Problem 2.116
The container shown in the figure below has circular cross sections. Find the vertical force
on the inclined surface. Also find the net vertical force on the bottom, EF. Is the vertical
force equal to the weight of the water in the container?
2′
2′
B
C
1′
A
D
Water
4′
E
F
4′
Solution 2.116
The vertical force on the inclined surface is equal to the weight of the water “above” it.
This “water volume” is
V = V cyl + V hole − V frustrum
ri
(
)
1
V = π ro2 − π ri2 ( − h ) − π h ro2 + ri2 + rori .
3
V = π ( 2 ft ) ( 3ft ) − π (1ft ) ( 3 − 1) ft
2
2
1
2
2
− π (1ft ) ( 2 ft ) + (1ft ) + ( 2 ft )(1ft )  = 24.1ft3


3
The vertical force FVi is
ri = 1", ro = 2", h = 1", = 3".
(
)
lb 

FVi = γ V =  62.4 3  24.1ft3 = FVi = 1500 lb
ft 

The pressure is uniform over the bottom EF so
lb 

2
FVb = pA = γ hA =  62.4 3  ( 7 ft ) π ( 2 ft )
ft 

or
FVb = 5490 lb
This force FVb is not equal
*CRC Standard
to the weight of water
Math Tables
in the container.
ℓ
h
ro
.
Problem 2.117
The 18-ft-long lightweight gate of the figure below is a quarter circle and is hinged at H .
Determine the horizontal force, P, required to hold the gate in place. Neglect friction at the
hinge and the weight of the gate.
P
6 ft
Water
Gate
H
Hinge
Solution 2.117
x1
P
FV
W
F1
FH
y1
FV
Hy
O
FH
Hx
For equilibrium (from free-body-diagram of fluid mass),
 Fx = 0
So that
Similarly,
FH = F1 = γ hc1 A1
lb   6 ft 

=  62.4 3  
 ( 6 ft ×18ft ) = 20200 lb
ft   2 

 Fy = 0
So that
lb   π

2

FV = W = γ H 2O × ( volume of fluid ) =  62.4 3   ( 6 ft ) ×18ft  = 31800 lb
ft   4


Also, x1 =
4 ( 6 ft ) 8
= ft
3π
π
(see the figure below)
y1 =
and
6 ft
= 2 ft
3
π R2
A = –––––
4R
–––
3π
4R
–––
3π
c
4
Ixc = Iyc = 0.05488R4
x
R
Ixyc = –0.01647R4
y
For equilibrium (from free-body-diagram of gate)
 M0 = 0
So that
P ( 6 ft ) = FH ( y1 ) + FV ( x1 )
or
P=
( 20200 lb )( 2 ft ) + (31800 lb ) 
6 ft

ft 
 π  = 20200 lb
8
Problem 2.118
The air pressure in the top of the 2-liter pop bottle and the
figure below is 40 psi , and the pop depth is 10 in. The
1-in. diameter
pair = 40 psi
bottom of the bottle has an irregular shape with a diameter
of 4.3 in. (a) If the bottle cap has a diameter of 1 in. what is 12 in.
10 in.
the magnitude of the axial force required to hold the cap in
4.3-in. diameter
place? (b) Determine the force needed to secure the bottom
2 in. of the bottle to its cylindrical sides. For this
calculation assume the effect of the weight of the pop is
negligible. (c) By how much does the weight of the pop increase the pressure 2 in. above the
bottom? Assume the pop has the same specific weight as that of water.
Solution 2.118
(a)
lb  π 

2
Fcap = pair × Areacap =  40 2    (1in.) = 31.4 lb
 in.   4 
(b)
Fsides
F1
 Fvertical = 0
Fsides = F1 = (pressure @ 2 in. above bottom) × (Area)
lb  π 

2
=  40 2    ( 4.3 in.)
 in.   4 
= 581lb
(c)
p = pair + γ h




lb 
lb   8 
1

= 40 2 +  62.4 3   ft  
in. 
ft   12  
in.2 
 144 2 
ft 

lb
lb
= 40 2 + 0.289 2
in.
in.
Thus, the increase in pressure due to weight = 0.289 psi
(which is less that 1% of air pressure).
bottom
Problem 2.119
In drilling for oil in the Gulf of Mexico, some divers have to work at a depth of 1300 ft.
(a) Assume that seawater has a constant density of 64 lb/ft3 and compute the pressure at
this depth. The divers breathe a mixture of helium and oxygen stored in cylinders, as shown
in the figure below, at a pressure of 3000 psia . (b) Calculate the force, which trends to blow
the end cap off, that the weld must resist while the diver is using the cylinder at 1300 ft .
(c) After emptying a tank, a diver releases it. Will the tank rise or fall, and what is its initial
acceleration?
8.0 in.
diameter
Weld
30 in.
Steel
p = 489 lbm/ft3
Spherical
cap
1.0 in.
thickness
Solution 2.119
(a)
The hydrostatic pressure is
 ft 2 
 lb 
or
p = γ sw h =  64 3  (1300 ft ) 
 144 in.2 
 ft 


(b)
p = 578psig
The net horizontal force on the end caps is
FN = Fin − Fout = pin Ain − pout Aout
and
τ = wall stress=
FN
FN
=
Awall Aout − Ain
2
2
− pout Dout
pin Ain − pout Aout pin Din
=
=
2
2
Aout − Ain
− Din
Dout
lb 
lb
2
2

 3000 2  ( 6 in.) − (14.7 − 578 ) 2 ( 8 in.)
in. 
in.
=
2
2
(8 in.) − ( 6 in.)
and
τ = 2500 psi.
(c)
The net vertical force on an empty tank and Newton's second law give
+ ↑ Fvert = FBuoy −W = ma
or
a=
FBuoy −W
=
m
FBuoy
m
−g
where m is the mass of the tank. Now
 π 

π 
FBuoy = γ sw V = γ sw   Dout 2 +   Dout3 
6
 4 

where
= 30 in. − 6 in. = 24 in. Also
 π 
m = ρsteel  
 4 
( Dout2 − Din2 ) +  π6  ( Dout3 − Din3 ) .
Substituting into the equation for a gives
a=
1
1

Dout 2 + Dout3 
6
4

1

Dout 2 − Din2 + Dout3 − Din3 
6

γ sw 
(
1
ρsteel 
4
) (
)
−g.
The numerical values give
( )
ft ⋅ lbm 
 lb   1
2 1 3 
3
 64 3   4 ( 24 ) 8 + 6 8  in.  32.2
2
ft
lb
sec


⋅



 − 32.2 ft
a=
lbm   1
1 3

sec 2
2
2
3 
3
 489 3   4 ( 24 ) 8 − 6 + 6 8 − 6  in.

ft  

(
or
a = −23.1
ft
sec 2
tank will fall
since a < 0.
) (
)
Problem 2.120
Hoover Dam is the highest arch-gravity type of dam in the United States. A cross section of
the dam is shown in the figure below (a). The walls of the canyon in which the dam is
located are sloped, and just upstream of the dam the vertical plane shown in the figure
below (b) approximately represents the cross section of the water acting on the dam. Use
this vertical cross section to estimate the resultant horizontal force of the water on the dam,
and show where this force acts.
45 ft
880 ft
727 ft
715 ft.
660 ft
290 ft
(a)
(b)
Solution 2.120
45 ft
880 ft
A1
A2
727 ft
A3
715 ft.
d
660 ft
(a)
290 ft
(b)
Break area into 3 parts as shown.
For area 1:
lb   1 

1
FR1 = γ hc A1 =  62.4 3    ( 715 ft )   ( 295 ft )( 715 ft )
ft   3 
2

= 1.57 × 109 lb
For area 3 : FR3 = FR1 = 1.57 ×109 lb
For area 2 :
lb   1 

FR2 = γ hc A2 =  62.4 3    ( 715 ft )( 290 ft )( 715 ft )
ft   2 

= 4.63 × 109 lb
Thus,
FR = FR1 + FR2 + FR3 = 1.57 × 109 lb + 4.63 × 109 lb + 1.57 ×109 lb
= 7.77 ×109 lb
Since the moment of the resultant force about the base of the dam must be equal to the
moments due to FR1 , FR2 , and FR3 it follows that
2
1
2
FR × d = FR1   ( 715 ft ) + FR2   ( 715 ft ) + FR3   ( 715 ft )
3
2
3
and
d=
(1.57 ×109 lb )  23  (715 ft ) + ( 4.63 ×109 lb )  12  (715 ft ) + (1.57 ×109 lb )  23  ( 715 ft )
7.77 ×109 lb
= 406 ft
Thus, the resultant horizontal force on the dam is 7.77 × 109 lb acting 406 ft up from the
base of the dam along the axis of symmetry of the area.
Problem 2.121
A plug in the bottom of a pressurized tank is conical in shape, as shown in the figure below.
The air pressure is 40 kPa, and the liquid in the tank has a specific weight of 27 kN/m3.
Determine the magnitude, direction, and line of action of the force exerted on the curved
surface of the cone within the tank due to the 40 − kPa pressure and the liquid.
40 kPa
Air
Liquid
3m
1m
60°
Solution 2.121
d
tan30 = 2
1
Pair A
𝒲
d = 2 tan30 = 1.155 m
2
π d 
volume of cone =   (1)
32
Fc
d
For equilibrium,
 Fvertical = 0
So that
Fc = pair A + aw
where Fc is the force the cone exerts of the fluid.
Also,
π 
pair A = ( 40 kPa )   d 2
4
( )
2
π 
= ( 40 kPa )   (1.155 m ) = 41.9 kN
4
And
2
π

π d 
 3 m 1m 
−
W = γ  d 2 ( 3 m ) −   (1m )  = γπ d 2 
32
12 
 4
 4


kN 

22
=  27 3  (π )(1.155 m )  m  = 75.4 kN
m 
3 

Thus,
Fc = 41.9 kN + 75.4 kN = 117 kN
And the force on the cone has a magnitude of 117 kN and is directed vertically downward
along the cone axis.
Problem 2.122
The homogeneous gate shown in the figure below consists of one quarter of a circular
cylinder and is used to maintain a water depth of4 m. That is, when the water depth
exceeds 4 m , the gate opens slightly and lets the water flow under it. Determine the weight
of the gate per meter of length.
Pivot
4m
1m
Solution 2.122
4R
–––
3π
Oy
ℓ2
O'x
y
ℓ1
x
W
gate
width = 1 m
ℓ3
FH
Water
W1
FV
ℓ4
Consider the free body diagram of the gate and a portion of the water as shown.
 Mo = 0, or
(1)
2W + 1W1 − FH 3 − FV 4 = 0, where
(2)
FH = γ hc A = 9.8 × 103
N
m3
(3.5 m )(1m )(1m ) = 34.3kN
since for the vertical side, hc = 4 m − 0.5 m = 3.5 m
Also,
(3)
FV = γ hc A = 9.8 ×103
N
m3
( 4 m )(1m )(1m ) = 39.2 kN
Also,
(4)
N  π
3
π
2
W1 = γ (1m ) − γ  (1m )  (1m ) = 9.8 × 103 3 1 −  m3 = 2.10 kN
m  4
4

(5)
Now,
(6)
1
3
1m
1m
(
)(
)
I xc
= 0.5 m + 12
= 0.524 m
3 = 0.5 m+ ( yR − yc ) = 0.5 m +
yc A
3.5 m (1m )(1m )
(7)
and
4 = 0.5 m and
2 = 1m −
4 (1m )
4R
= 1−
= 0.576 m
3π
3π
To determine 1, consider a unit square that consist of a quarter circle and the remainder as
show in the figure. The centroids of areas (1) and (2) are as indicated.
I
0.5
(2)
C2
C
I
4
–––
3π
C1
(1)
ℓ1
0.5 – ℓ1
Thus,
4 

 0.5 −
 A2 = ( 0.5 − 1 ) A1
3π 

So that with A2 =
π
4
(1)2 =
π
4
and A1 = 1 −
π
4
this gives
4 π

 π
 0.5 −
 = ( 0.5 − 1 ) 1 − 
3π  4

 4
or
(8)
1 = 0.223 m
Hence, by combining Eqs. (1) through (8):
( 0.576 m )W + ( 0.223 m )( 2.10 kN ) − ( 34.3kN )( 0.524 m ) − ( 39.2 kN )( 0.5 m ) = 0
or
W = 64.4 kN
Problem 2.123
The concrete (specific weight =150 lb/ft3) seawall of the figure below has a curved surface
and restrains seawater at a depth of 24 ft . The trace of the surface is a parabola as
illustrated. Determine the moment of the fluid force (per unit length) with respect to an axis
through the toe (point A).
Seawater
y = 0.2x2
24 ft
y
A
x
15 ft
Solution 2.123
Volume = V
xc
C
D
Seawater
y = 0.2x2
24 ft
F1
y1
y
𝒲
A
B
x
15 ft
The components of the fluid force acting on the wall are F1 and W as shown on the figure
where
lb   24 ft 

F1 = γ hc A =  64.0 3  
 ( 24 ft ×1ft )
ft   2 

= 18400 lb
and
y1 =
24 ft
= 8ft
3
Also,
W =γV
To determine V find area BCD.
y
Thus,
A= 
x0
0
dx
C
( 24 − y ) dx = 0 ( 24 − 0.2x2 ) dx
x0
dA
x
0

0.2 x3 
= 24 x −

3 

0
24
B
X0 = √120
(Note: All lengths in ft)
Thus,
)
lb 

W =  64.0 3  175 ft3 = 11200 lb
ft 

To locate centroid of A:
( 24 − y ) x dx = 0 ( 24x − 0.2x3 ) dx = 12x02 −
0
x0
xc A =  x dA = 
0
x0
x0
and
12
( 120 ) −
2
0.2
( 120 )
4
175
4
= 4.11ft
Thus,
M A = F1 y1 −W (15 − xc )
= (18400 lb )( 8ft ) − (11200 lb )(15 ft − 4.11ft )
= 25200 ft ⋅ lb
x
X0
V = A × 1ft = 175 ft3
(
24–y
x
And with x0 = 120 , A = 175 ft 2 so that
xc =
D
0.2 x04
4
Problem 2.124
A step-in viewing window having the shape of a half-cylinder is built into the side of a large
aquarium. See the figure below. Find the magnitude, direction, and location of the net
horizontal forces on the viewing window.
x
z
y
25′
p = 64 lbm/ft3
R = 5′
10′
2′
Steps up to
window
Solution 2.124
Due to symmetry, the net force parallel to the wall is zero or
Fz = 0
The net horizontal force perpendicular to the wall is
 lb 
Fx = γ hc A =  64 3  ( 25 + 5 ) ft (10 ft ×10 ft )
 ft 
Fx = 1.92 × 105 lb
The vertical location of Fx is
1 3
bh
I
(10 ft )2 or
h2
y p = yc + xc = yc + 12
= yc +
= 30 ft +
yc A
yc bh
12 yc
12 ( 30 ft )
The net horizontal force also acts through the coordinate
z = 0 and acts in
an outward direction.
y p = 30.3ft
Problem 2.125
Find the magnitude, direction, and location of the net vertical force acting on the viewing
window in Problem 2.124.
x
z
y
25′
p = 64 lbm/ft3
R = 5′
10′
2′
Steps up to
window
Solution 2.125
The net vertical force must equal the weight of fluid inside the viewing window. Then
Fy = 25100 lb,
2
π
  lb 
π 
Fy = γ V = γ h  R 2  =  64 3  (10 ft )   ( 5 ft ) or
acting upword.
2
  ft 
2
This net vertical force acts through the centroid of the window volume. Using Appendix B
gives
x=
4R 4 ( 5 ft )
=
3π
3π
or
x = 2.12 ft
Problem 2.126
A l0-m-long log is stuck against a dam, as shown in the figure below. Find the magnitudes
and locations of both the horizontal force and the vertical force of the water on the log in
terms of the diameter D. The center of the log is at the same elevation as the top of the dam.
D
Water
Solution 2.126
FH
Consider the water forces on the log as shown on the right.
The horizontal forces FH is on the top portion only and is
5
γ
D
 D  D 
FH = γ    
 4  2 
where
=
0.
m
FV
is the log length. Assuming 10 C water, Table A.5 gives
kg  
m

FH =  1000 3  9.81 2  ( 0.25 m )( 0.5 m )(10 m ) = 12300 N = FH

m 
s 
The location of FH is
2  D  2  0.5 
m  = 0.167 m = y p
yp =   = 
3 2  3 2

The vertical force FV is the weight of water “above” the bottom of the log minus the weight
of water above the top half of the log. This is
 π D2  D 
 D 2 π D 2   γ D3  3π

FV = γ 
+  D −
−
=
+ 1



 4
16  
4  4
2

 8

kg  
m
2

1000 3  9.81 2  (10 m )(1.0 m ) 3π


m 
s 
=
+ 1

4
 4

FV = 82300 N
The location x of FV is found by first locating the centroid of area A1 by
x1 =
A1+2 x1+2 − A2 x2
.
A1
x2
Using Table B
x1
2
 D   D   π D  D 2D 
 −
    − 

 2   4   16   2 3π 
2
x1 =
2
x
1
2
 D  πD
  −
16
2
 1 π  1 2 
 16 − 16  2 − 3π  

D
=
π
1


−


4 16
= 0.112 D
2
D
y
FV1
and is the location of FV 1. The location of
FV 2 is
x
D
. The location of FV is
2
D
FV 2   − FV 1 ( 0.112 D )
2
x=
FV
 π D2 D2   D   D2 π D2 
+
−

 − 
 ( 0.112 D )
2   2   4
16 
 B

=
π D2 D2  D2 π D2 
+
−
−

B
2  4
16 
1π

 1  π 
 + 1 − 0.112   1 − 
4 4 
 4  4  D
= 
1π
 1 π 
 + 1 − 1 − 
2 4  4 4 
x = 0.525 D
y
FV2
weight of water “above”
bottom portion of log.
weight of water above top left
portion of log.
Problem 2.127
Find the net horizontal force on the 4.0-m -long log shown in the figure below.
1.0 m
D = 1.0 m
0.5 m
Solution 2.127
The force FL on the left side of the log is the horizontal force on
FL
D
the horizontally projected area of the log. This horizontally
projected area measures D = 1.0 m by 4.0 m and gives
FL = ρ ghc A
 N ⋅ s2 
kg  
m

= 1000 3   9.81 2  ( 0.5 m )(1.0 m × 4.0 m ) 
 kg ⋅ m 

m 
s 


= 19600 N = 19.6 kN
The force FR on the right side of the log is the horizontal force on the horizontally
projected area of the lower half of the log. This horizontally projected area measures
D
= 0.5m by 4.0 m and gives
2
 N ⋅ s2 
kg  
m

FR = ρ ghc A = 1000 3   9.81 2  ( 0.25 m )( 0.5 m × 4.0 m ) 
 kg ⋅ m 
m 
s 



= 4910 N = 4.91 kN
The net horizontal force is
F = FL − FR = 19.6 kN − 4.91kN
F = 14.7 kN, acting to right.
FR
Problem 2.128
An open tank containing water has a bulge in its vertical side that is semicircular in shape
as shown in the figure below. Determine the horizontal and vertical components of the
force that the water exerts on the bulge. Base your analysis on a 1-ft length of the bulge.
6 ft
Bulge
Water
3 ft
Solution 2.128
FH ~ horizontal force of wall on fluid
𝒲
F1
FH
FV ~ vertical force of wall on fluid
W = γ H 2OVvol
FV
2
lb   π ( 3ft ) 

 (1ft )
=  62.4 3  

2
ft  


= 882 lb
lb 

F1 = γ hc A =  62.4 3  ( 6 ft + 3ft )( 6 ft ×1ft ) = 3370 lb
ft 

For equilibrium,
FV = W = 882 lb ↑
and
FH = F1 = 3370 lb ←
The force the water exerts on the bulge is equal to, but opposite in direction to FV and FH
above. Thus,
( FH )wall = 3370 lb →
( FV )wall = 882 lb ↓
Problem 2.129
A closed tank is filled with water and has a 4-ft-diameter hemispherical dome as shown in
the figure below. A U-tube manometer is connected to the tank. Determine the vertical
force of the water on the dome if the differential manometer reading is 7 ft and the air
pressure at the upper end of the manometer is 12.6 psi.
pA
Air
4-ft diameter
20 psi
5 ft
2 ft
Water
2 ft
Gage
fluid
(SG = 3.0)
Solution 2.129
FD
For equilibrium,
 Fvertical = 0
𝒲
So that
FD = pA −W
pA
Eq. (1)
Where FD force exerted by dome on the fluid and p is water pressure at the dome base.
From the manometer,
pA + γ gf ( 7 ft ) − γ H2O ( 4 ft ) = p
lb  
in.2 
lb 
lb 



p = 12.6 2  144 2  + ( 3.0 )  62.4 3  ( 7 ft ) −  62.4 3  ( 4 ft )


in.  
ft 
ft 
ft 



= 2880
lb
ft 2
Thus, from Eq. (1) with volume of sphere =
π
6
( diameter )3
lb   π 
lb 

2 1 π
3
FD =  2880 2    ( 4 ft ) −  ( 4 ft )   62.4 3  = 35100 lb
2 6
ft   4 
ft 


The vertical force that the water exerts on the dome is 35100 lb ↑.
Problem 2.130
A 3-m-diameter open cylindrical tank contains water and has a hemispherical bottom as
shown in the figure below. Determine the magnitude, line of action, and direction of the
force of the water on the curved bottom.
Water
8m
3m
Solution 2.130
Force = weight of water supported by hemispherical bottom
= γ H2 O ( volume of cylinder ) − ( volume of hemisphere ) 
= 9.80
kN  π
π
2
3
m
8
m
−
(
)
(
)
( 3 m )3 

3 4
12
m 

= 485 kN
The force is directed vertically downward, and due to symmetry it acts on the hemisphere
along the vertical axis of the cylinder.
485 kN
Problem 2.131
Three gates of negligible weight are used to hold back water in a channel of width b as
shown in the figure below. The force of the gate against the block for gate (b) is R.
Determine (in terms of R) the force against the blocks for the other two gates.
h
2
h
2
Hinge
h
h
h
Block
(a)
(b)
(c)
Hy
Solution 2.131
Hx
For case (b)
yR
γh b
h
FR = γ hc A = γ   ( h × b ) =
2
2
2
FR
and
FB = R
yR =
2
h
3
Thus,
 MH = 0
So that
2
γ h2 b
2 
 2  γ h b 
→ R=
hR =  h  FR =  h  

3
3 
 3   2 
R=
γ h2 b
3
(1)
For case (a) on free-body-diagram shown
FR =
γ h2 b
2
4h
(see the figure below)
3π 2
Hy H
x
(from above) and
𝒲
yR
2
yR = h
3
FR
and
  h 2

π  

πγ h2 b
2



W =γ ×V o =γ
( b ) =
 4
16




FB
(gate
width = b)
π R2
A = –––––
Thus,  M H = 0
2
Ixc = 0.1098R4
 h 4h 
2 
So that W  −
 + FR  h  = FB h
 2 6π 
3 
πγ h2 b  h
4h  γ h2 b  2 
 −
+
 h  = FB h
16  2 6π 
2 3 
FB = γ h2 b(0.390) = 3R(0.390) →
c
y
R
x
Iyc = 0.3927R4
4R
–––
3π
Ixyc = 0
R
FB = 1.17R
For case (c), for the free-body-diagram shown, the force FR1 on the curved section passes
through the hinge and therefore does not contribute to the moment around H. On bottom
part of gate
 3h   h  3
FR2 = γ hc A = γ    × b  = γ h2 b
 4  2  8
Hy
and
Hx
3
1
h
b)  
(
I
12
 2  + 3h
yR2 = xc + yc =
yc A
 3h   h  4
  ×b
 4  2 
28
= h
36
Thus,
 MH = 0
So that
 28 
FR2  h  = FB h
 36 
or
3
 28  7
FB =  γ h2 b   = γ h2 b
8
 36  24
From Eq. (1) γ h2 b = 3R , thus
7
FB = R = 0.875R
8
yR
2
FR
FR
FB
2
1
Problem 2.133
An iceberg (specific gravity 0.917 ) floats in the ocean (specific gravity 1.025 ). What percent
of the volume of the iceberg is under water?
Solution 2.133
SG = 0.917
ice
SG = 1.025
ocean
For equilibrium,
W = weight of iceberg = FB = buoyant force
or
V iceγ ice = V subγ ocean , where V sub = volume of ice submerged.
Thus,
γ
SGice
V sub
0.917
= ice =
=
= 0.895 = 89.5%
V ice γ ocean SGocean 1.025
Problem 2.134
A floating 40-in.- thick piece of ice sinks 1 in. with a 500-lb polar bear in the center of the
ice. What is the area of the ice in the plane of the water level? For seawater, S = 1.03.
Solution 2.134
Without the polar bear on the ice, the submerged depth d of the ice is found by equating
the weight of the ice and the buoyant force. Denoting the pure water specific weight by
γ and the ice area by A gives
FB = Wice
area A
d
or
t
Wice = γ SAd .
The ice sinks an additional depth d ′ with the bear in the center of the ice. Equating the new
buoyant force to the weight of the ice plus bear gives
FB = Wice + Wbear ,
γ SA ( d + d ′ ) = γ SAd +Wbear,
or
A=
Wbear
500 lb
or A = 93.4 ft 2
=
lb 
γ Sd ′ 
1 
 62.4 3  (1.03 )  12 ft 
ft 



Problem 2.135
A spherical balloon filled with helium at 40 °F and 20 psia has a 25 − ft diameter. What
load can it support in atmospheric air at 40 °F and 14.696 psia ? Neglect the balloon’s
weight.
Solution 2.135
For static equilibrium, the buoyant force must equal the load. Neglecting the weight of the
balloon and assuming air and helium to be ideal gases, the load is
L = FB = ( γ air − γ He ) V = ( ρair − ρ He ) gV
 p 
 p    g  4

=   −       π R 3 

 R air  R He   T   3
Using Table A.4, the numerical values give

ft   4π 
3
lb
lb  
( 20 ×144 ) 2   32.2 2   3  (12.5 ft )
 (14.696 ×144 ) 2

sec  
ft −
ft  
L=
  53.35 ft ⋅ lb   386 ft ⋅ lb   500 o R  32.2 ft ⋅ lbm 
 



 
lbm ⋅ o R  
lbm ⋅ o R  
 lb ⋅ sec 2 
(
or
L = 527 lb
)
Problem 2.136
A river barge, whose cross section is approximately rectangular, carries a load of grain. The
barge is 28 ft wide and 90 ft long. When unloaded, its draft (depth of submergence) is 5 ft
and with the load of grain the draft is 7 ft . Determine: (a) the unloaded weight of the
barge, and (b) the weight of the grain.
Solution 2.136
(a)
For equilibrium,
Wb
 Fvertical = 0
5 ft
So that
FB
Wb = FB = γ H2O × (submerged volume)
Wb ~ weight of barge
(unloaded)
lb
) ( 5 ft × 28ft × 90 ft )
ft3
= 786000 lb
= (62.4
(b)
 Fvertical = 0
Wb + W g
WB +Wg = FB = γ H 2O × (submerged volume)
7 ft
lb
) ( 7 ft × 28ft × 90 ft ) − 786,000 lb
ft3
= 315000 lb
Wg = (62.4
FB
Wg ~ weight of grain
Problem 2.137
A barge is 40 ft wide by 120 ft long. The weight of the barge and its cargo is denoted by W .
(
)
When in salt-free riverwater, it floats 0.25 ft deeper than when in seawater γ = 64 lb/ft3 .
Find the weight W .
Solution 2.137
In both cases, the weight W must equal the
weight of the displaced water or
W = γ SFW A ( d + 0.25 ft )
= γ SW Ad
d + 0.25′
γ SFW
A
salt-free water
Soling for d gives
γ SW Ad = γ SFW A ( d + 0.25 ft )
or
d
γ SW
( 0.25 ft ) γ SFW =
d=
γ SW − γ SFW
( 0.25 ft )  62.4 3 
lb
ft  = 9.75 ft .

lb
( 64.0 − 62.4 ) 3
ft
Then
lb 

W = γ SW Ad =  64.0 3  ( 40 ×120 ) ft 2 ( 9.75 ft )
ft 

 short ton 
W = 3.00 ×106 lb 
,
 2000 lb 
or
W = 1500 short tons.
sea water
Problem 2.138
When the Tucurui Dam was constructed in northern Brazil, the lake that was created
covered a large forest of valuable hardwood trees. It was found that even after 15 years
underwater, the trees were perfectly preserved and underwater logging was started. During
the logging process, a tree is selected, trimmed, and anchored with ropes to prevent it from
shooting to the surface like a missile when cut. Assume that a typical large tree can be
approximated as a truncated cone with a base diameter of 8ft , a top diameter of 2 ft , and a
height of 100 ft. Determine the resultant vertical force that the ropes must resist when the
completely submerged tree is cut. The specific gravity of the wood is approximately 0.6 .
Solution 2.138
W ~ weight , FB ~ buoyant force , T ~ tension in ropes
For equilibrium,
 Fvertical = 0
𝒲
FB
T = FB −W
For a truncated cone,
πh
(
r12 + r1r2 + r22
3
where: r1 = base radius
Volume =
)
r2 = top radius
h = height
Thus,
V tree =
(π )(100 ft )  4 ft 2 + 4 ft ×1ft + 1ft 2  = 2200 ft3
( ) (
) ( ) 

3
For buoyant force,
FB = γ H2O × V tree = (62.4
lb
ft
3
(
)
) 2200 ft3 = 137000 lb
For weight,
(
)
lb 

W = γ tree × V tree = ( 0.6 )  62.4 3  2200 ft3 = 82400 lb
ft 

Therefore,
T = 137000 lb − 82400 lb = 54600 lb
T
Problem 2.140
An inverted test tube partially filled with air floats in a plastic water-filled soft drink bottle
as shown in the figure below. The amount of air in the tube has been adjusted so that it just
floats. The bottle cap is securely fastened. A slight squeezing of the plastic bottle will cause
the test tube to sink to the bottom of the bottle. Explain this phenomenon.
Air
Test tube
Water
Plastic bottle
Solution 2.140
p
FB
po
h
Volume
displaced
𝒲
where the test tube is floating the weight of the tube, W , is balanced by the buoyant force,
FB , as shown in the figure. The buoyant force is due to the displaced volume of water as
shown. This displaced volume is due to the air pressure, p, trapped in the tube where
p = po + γ H 2O h. When the bottle is squeezed, the air pressure in the bottle, po, is increased
slightly and this in turn increases p, the pressure compressing the air in the test tube. Thus,
the displaced volume is decreased with a subsequent decrease in FB . Since W is constant, a
decrease in FB will cause the test tube to sink.
Problem 2.141
A child’s balloon is a sphere 1 ft. in diameter. The balloon is filled with helium
( ρ = 0.014 lbm/ft3 ). The balloon material weighs 0.008 lbf/ft 2 of surface area. If the child
releases the balloon, how high will it rise in the Standard Atmosphere. (Neglect expansion
of the balloon as it rises.)
Solution 2.141
ρ airgV
A force balance in the vertical direction for the balloon gives
g
D
He
+ ↑  Fz = 0 = ρair gV − ρ He gV − wA
for the balloon at rest at its highest elevation. Then
ρ HegV + wA
w = wt/surface area
A = surface area
ρair =
ρ He gV + wA
gV
(
)
w πD
wA
= ρ He +
= ρ He +
gV
 π D3 
g
 6 


2
lb  
ft ⋅ lbm 

6  0.008 2   32.2

6w
lbm
ft  
lb ⋅ sec 2 
= ρ He +
= 0.014 3 + 
ft 
gD

ft
 32.2 2  (1.0 ft )
sec 

lbm
= 0.062 3 .
ft
Interpolating Table A.2 for the Standard Atmosphere,
 0.06590 − 0.062 
z = elevation = 5000 ft + 5000 ft 
 →
 0.06590 − 0.05648 
z = 7,070 ft
Problem 2.142
A 1-ft-diameter, 2-ft-long cylinder floats in an open tank containing a liquid having a
specific weight . A U-tube manometer is connected to the tank as shown in the figure
below. When the pressure in pipe A is 0.1 psi below atmospheric pressure, the various fluid
levels are as shown. Determine the weight of the cylinder. Note that the top of the cylinder
is flush with the fluid surface.
pa = – 0.1 psi
1 ft
A
1 ft
Water
2 ft
2 ft
Cylinder
Sp. wt. = γ
Gage fluid SG = 1.5
0.5 ft
Solution 2.142
From a free-body-diagram of the cylinder
𝒲
 Fvertical = 0
So that
FB
2
π 
W = FB = γ   (1ft ) ( 2 ft )
4
=
(1)
πγ
2
A manometer equation gives,
(
)
γ ( 3.5 ft ) − ( SG ) γ H O ( 2.5 ft ) − γ H O (1ft ) = pA


2
γ ( 3.5 ft ) − (1.5 )  62.4
γ = 80.6
2
lb 
lb 
lb  
in.2 


2.5
ft
62.4
1ft
0.1
144
−
=
−
(
)
(
)







ft3 
ft3 
in.2  
ft 2 


lb
ft3
lb 
π

Therefore, W =  ft3   80.6 3  = 127 lb
ft 
2

Problem 2.143
A not-too-honest citizen is thinking of making bogus gold bars by first making a hollow
iridium ( SG = 22.5 ) ingot and plating it with a thin layer of gold ( SG = 19.3 ) of negligible
weight and volume. The bogus bar is to have a mass of 100 lbm. What must be the volumes
of the bogus bar and of the air space inside the iridium so that an inspector would conclude
it was real gold after weighing it in air and water to determine its density? Could lead
( SG = 11.35 ) or platinum ( SG = 21.45 ) be used instead of iridium? Would either be a good
idea?
Solution 2.143
SGx = 22.5 ( iridium ) ;
SGG = 19.3 ( gold ) ;
V BB = V x + V AS ;
mBB = mx = 100 lbm
Neglect the weight of air in the air space and the buoyant force of air on the bar.
The volume of a pure gold bar would be
V GB =
WGB
γG
.
The bogus bar must have the same volume and weight as the pure gold bar so it will weigh
like a solid gold bar in water. The volume condition gives
V GB = V BB = V AS + Vx.
Since WGB = Wx ,
V AS + V x = V GB =
WGB
γG
=
Wx
γG
, V AS + V x =
γxV x
,
γG
γ


S
V AS = V x  x − 1 = V x  x − 1 .
 γG

 SG

The numerical value of the iridium volume is
Vx=
Wx
γx
=
WGB
γx
=
100 lb
lb 

 22.5 × 62.4 3 
ft 

= 0.0712 ft3.
 22.5 
The air space volume is V AS = 0.0712 ft3 
− 1 or
 19.3 
V AS = 0.0118ft3 .
The bogus bar volume is
V BB = V AS + V x = ( 0.0118 + 0.0712 ) ft3
or
V BB = 0.0830 ft3 .
Lead will not work because it is less dense than gold.
Platinum could work because it is more dense than gold.
However, platinum is more expensive per unit weight than
gold, so it would be a foolish choice.
Problem 2.144
A solid cylindrical pine ( SG = 0.50 ) spar buoy has a cylindrical
2′
lead ( SG = 11.3 ) weight attached, as shown in the figure below.
Determine the equilibrium position of the spar buoy in seawater
(i.e., find d ). Is this spar buoy stable or unstable? For seawater,
SG = 1.03.
16′
d
Lead
0.5′
A
Solution 2.144
The equilibrium position is found by equating the buoyant force and
the body weight.
FB = W
γ sw dA = γ
d=
=
A + γ p pA
γ
+γ p p
γ sw
SG
=
ℓp
+ SG p p
d
ℓcp
SGsw
11.3 ( 0.5 ft ) + 0.50 (16 ft )
1.03
→
ℓc
d = 13.3ft
ℓcℓ
ℓℓ
Since d < 13.8ft (the total length of the spar buoy), the spar buoy
floats. We now have to check the stability of the buoy.
I=
π
4
π
( radius )4 = (1ft )4 = 0.7854 ft 4 ,
4
c = distance from bottom of buoy to center of gravity of buoy
c =
cW
+ cpWcp
W +Wp
=
 
 2  (γ A
 

)+
p
(
 γ pA p
2 

γ A + γ pA p
SG
=
+
)

 
p
 2  + SG p p  + 2 
 

 = 11.3 ( 0.5 )( 0.25 ) + 0.5 (16 )( 0.5 + 8 ) ft
SG + SG p p
11.3 ( 0.5 ) + 0.5 (16 )
c = 5.09 ft
d 13.3ft
=
= 6.65 ft , n =
2
2
c−
d
= 5.09 ft − 6.65 ft = −1.56 ft ,
2
m=
V s = Ad = π (1ft ) (13.3ft ) = 41.8ft3,
2
0.7854 ft 4
I
−n =
− ( −1.56 ft ) = 1.58ft
Vs
41.8ft3
m>0 →
buoy is stable
Problem 2.145
When a hydrometer (see the figure below) having a stem diameter
of 0.30 in. is placed in water, the stem protrudes 3.15 in. above the
Fluid
surface
Hydrometer
water surface. If the water is replaced with a liquid having a
specific gravity of 1.10, how much of the stem would protrude
above the liquid surface? The hydrometer weighs0.042 lb.
Solution 2.145
When the hydrometer is floating its weight, W , is balanced by the
buoyant force, FB .,
 Fvertical = FB −W = 0
(γ H O ) ∀1 = W = ( SG ) (γ H O ) ∀2
2
∀2 =
2
∀1
SG
FB
For water,
W
0.042 lb
V1 =
=
= 6.73 × 10 −4 ft3
γ H 2O 62.4 lb
ft3
𝒲
For other liquid,
V2 =
6.73 ×10 −4 ft3
= 6.12 × 10 −4 ft3
1.10
Therefore,
V1 − V2 = ( 6.73 − 6.12 ) ×10 −4 ft3 = 0.61×10 −4 ft3
The change in submergence depth occurs with only the stem protruding from the surface.
in.3 
2
π 
−4 3 
0.30
in.
0.61
10
ft
1728
Δ
=
×
(
)


 

ft3 
4

(
)
Δ = 1.49 in.
With the new liquid the stem would protrude 3.15 in.+1.49 in. = 4.64 in. above surface
Problem 2.146
A 2-ft-thick block constructed of wood ( SG = 0.6 ) is
submerged in oil ( SG = 0.8 ) and has a 2-ft-thick
6 ft
Oil
3
aluminum (specific weight = 168 lb / ft ) plate
attached to the bottom as indicated in the figure
below. Determine completely the force required to
hold the block in the position shown. Locate the
force with respect to point A.
4 ft
0.5 ft
Aluminum
A
10 ft
Solution 2.146
10 ft
3
Equilibrium:  Fvertical = F −Ww + FBw −Wa + FBa = 0
(
Ww
)
Ww = ( SGw ) γ H 2O Vw
lb   1 

= ( 0.6 )  62.4 3    (10 ft × 4 ft × 2 ft ) = 1500 lb
ft   2 

FBw
A
lb 

Wa = 168 3  ( 0.5 ft ×10 ft × 2 ft ) = 1680 lb
ft 

(
Wa
)
FBw = ( SGoil ) γ H 2O Vw
ℓ
F FBa
5 ft
w ~ wood
a ~ aluminum
F ~ force to hold block
lb  1 

= ( 0.8 )  62.4 3    (10 ft × 4 ft × 2 ft ) = 2000 lb
ft   2 

lb 

FBa = ( SGoil ) γ H 2O Va = ( 0.8 )  62.4 3  ( 0.5 ft ×10 ft × 2 ft ) = 499 lb
ft 

(
)
Therefore, F = 1500 lb − 2000 lb +1680 lb − 499 lb → F = 681 lb upward
Equilibrium:
 10 
F =  ft  (Ww − FBw ) + ( 5 ft ) (Wa − FBa )
3 
10
( 681lb ) =  ft  (1500 lb − 2000 lb ) + (5 ft )(1680 lb − 499 lb ) = 6.22 ft
3 
 Ma = 0 →
F acts 6.22 ft to the right of point A
Problem 2.147
How much extra water does a 147 − lb concrete canoe displace compared to an
ultralightweight 38 − lb Kevlar canoe of the same size carrying the same load?
Solution 2.147
For equilibrium,
𝒲
 Fvertical = 0
FB
and
W = FB = γ H 2O V and V is displaced volume.
For concrete canoe,
lb 

147 lb =  62.4 3  Vc
ft 

Vc = 2.36 ft3
For Kevlar canoe,
lb 

38 lb =  62.4 3  Vk
ft 

Vk = 0.609 ft3
Extra water displacement = 2.36 ft3 − 0.609 ft3
= 1.75 ft3
Problem 2.148
A submarine is modeled as a cylinder with a length of 300 ft, a diameter of 50 ft, and a
conning tower as shown in the figure below. The submarine can dive a distance of 50 ft
from the floating position in about 30 s. Diving is
3% of cylinder
7% of cylinder volume
volume
accomplished by taking water into the ballast
tank so the submarine will sink. When the
pwater = 64 lbm/ft3
submarine reaches the desired depth, some of the
3′
water in the ballast tank is discharged leaving the
25′
submarine in “neutral buoyancy” (i.e., it will
50′
neither rise nor sink). For the conditions
illustrated, find (a) the weight of the submarine
Ballast tank
and (b) the volume (or mass) of the water that
Water
must be in the ballast tank when the submarine is
Partially
Totally
submerged
submerged
in neutral buoyancy. For seawater, S = 1.03.
position
position
Solution 2.148
(a) Denoting the cylinder radius by R, the submarine weight is equal to the buoyant
force so
W = FB = γ V submerged
(
= γ π R2
) (1.03)
when the submarine is in the partially submerged position. The numerical values
give
 lb 
2
W =  64 3  π ( 25 ft ) ( 300 ft )(1.03 ) or
 ft 
W = 3.88 ×107 lb
(b) For neutral buoyancy at the lower depth, the submarine weight W plus the ballast
weight WB must equal the buoyant force so
(
W + WB = FB = γ π R 2
or
(
WB = γ π R 2
) (1.10 )
) (1.10 ) −W .
The ballast volume V B =
(
V B = π R2
WB
γ
so
7
× 10 lb
) (1.10 ) − Wγ = π ( 25 ft )2 (300 ft )(1.10 ) − 3.88
 lb 
V B = 41700 ft3
 64 3 
 ft 
Problem 2.150
When an automobile brakes, the fuel gage indicates a fuller tank than when the automobile
is traveling at a constant speed on a level road. Is the sensor for the fuel gage located near
the front or rear of the fuel tank? Assume a constant deceleration.
Solution 2.150
accelerating
automobile
so
sensor located
in front of
fuel tank.
decelerating
( braking )
automobile
Problem 2.151
An open container of oil rests on the flatbed of a truck that is traveling along a horizontal
road at 55 mi / hr. As the truck slows uniformly to a complete stop in 5 s, what will be the
slope of the oil surface during the period of constant deceleration?
Solution 2.151
dy
z
dz
y
x
slope =
ay =
ay
dz
=−
dy
g + az
final velocity – initial velocity
time interval
m 



0 − ( 55 mph )  0.4470 s 
mph 


 = −4.92 m
=
5s
s2
Thus,
m

−4.92 2 

dz
s  = 0.502
=−
m
dy
9.81 2 + 0
s
Problem 2.152
A 5-gal, cylindrical open container with a bottom area of 120 in.2 is filled with glycerin and
rests on the floor of an elevator. (a) Determine the fluid pressure at the bottom of the
container when the elevator has an upward acceleration of 3 ft / s 2 . (b) What resultant force
does the container exert on the floor of the elevator during this acceleration? The weight of
the container is negligible. (Note: 1 gal = 231 in.3 )
Solution 2.152
hA = volume
 231in.3 
h 120 in.2 = ( 5 gal ) 
 gal 


(
)
h = 9.63 in.
h
ft
s
az = 3 ––––
2
z
pb
∂p
= − ρ ( g + az )
∂z
(a)
area = A
Thus,
pb
0
0
h
 dp = − ρ ( g + az )  dz
and
pb = ρ ( g + az ) h
slugs  
ft
ft  9.63 

ft 
=  2.44 3   32.2 2 + 3 2  
ft  
s
s   12 

lb
= 68.9 2
ft
(b)
From free-body-diagram of container,
F f = pb A
 1ft 2 
lb 

=  68.9 2  120 in.2 
 144 in.2 
ft 



(
)
= 57.4 lb
Thus, force of container on floor is 57.4 lb downward .
pb A
Ff
Problem 2.153
A plastic glass has a square cross section measuring 2½ in. on a side and is filled to within
½ in. of the top with water. The glass is placed in a level spot in a car with two opposite
sides parallel to the direction of travel. How fast can the driver of the car accelerate along a
level road without spilling any of the water?
Solution 2.153
2.5″
1 ″
––––
2
Slope of water surface
a
= − car
g
or
water
acar = − g ( slope )
ft   1.0 in. 

= −  32.2 2   −

sec   2.5 in. 

or
acar = 12.9
ft
sec 2
z
x
ax
Problem 2.154
The cylinder in the figure below accelerates to the left at the
rate of 9.80 m/s 2 . Find the tension in the string connecting
Sfluid = 1.0
g
Srod = 2.0
String
at rod of circular cross section to the cylinder. The volume
between the rod and the cylinder is completely filled with
water at 10 °C.
1.0 cm
8.0 cm
acyl = 9.8 m/s2
Solution 2.154
First find the pressure difference in the water over a length
= 8.0 cm . Since gravity is perpendicular to the rod, Eq.(2.41)
gives
Sfluid = 1.0
g
Srod = 2.0
String
1.0 cm
8.0 cm
dp = − ρ ax d x
acyl = 9.8 m/s2
For the x-direction. Integrating gives
ax = 9.80 m/s2
p2 − p1 = − ρ ax ( x2 − x1 ) .
p2
p1
For 10 °C water, Table A.5 gives
x1
 m 
kg  
m
N

p2 − p1 = − 1000 3   9.80 2  ( 8.0 cm ) 
 = −784 2

m 
s 
m
 100 cm 
x2
We next apply Newton’s second law to the rod
←
T
+  Fx = max,
m
p2A
T + ( p1 − p2 ) A = ma x ,
x
T = ( p2 − p1 ) A + ma x .
Using the specified information,
 π D2 
m = ρw Srod A = ρw Srod 
 4 


3
 1m 
kg 

π 
2
= 1000 3  ( 2.0 )   (1.0 cm ) ( 8.0 cm ) 
 = 0.0126 kg
m 

4
 100 cm 
A=
π D2
4
2
2
m 
−5 2
= (1.0 cm ) 
 = 7.854 ×10 m .
4
100
cm


π
Therefore,
(
)
N 
m


T =  −784 2  7.854 ×10 −5 m 2 + ( 0.0126 kg )  9.80 2 
m 
s 


T = 0.062 N
p1A
Problem 2.155
A closed cylindrical tank that is 8ft in diameter and 24 ft long is completely filled with
gasoline. The tank, with its long axis horizontal, is pulled by a truck along a horizontal
surface. Determine the pressure difference between the ends (along the long axis of the
tank) when the truck undergoes an acceleration of 5 ft / s 2 .
Solution 2.155
24 ft
z
x
(1)
y
(2)
ay = 5
ft
s2
∂p
= − ρ ay
∂y
Thus,
p2
24
p1
0
 dp = − ρ ay  dy
Where p = p1 at y = 0 and p = p2 at y = 24 ft,
and
p2 − p1 = − ρ ay ( 24 ft )
slugs   ft 

= − 1.32 3   5 2  ( 24 ft )
ft   s 

lb
= −158 2
ft
or
p1 − p2 = 158
lb
ft 2
Problem 2.156
The cart shown in the figure below
measures 10.0 cm long and 6.0 cm
high and has rectangular cross
sections. It is half-filled with water
and accelerates down a 20° incline
plane at a = 1.0 m/s2 . Find the
height h .
ℓ = 10.0 cm
y
g
h
z
a
20°
x
x
Solution 2.156
Unfortunately, there are 2 x-directions in the problem statement.
Noting that the gravisty vector is in the negative z-direction, change the label on the axis
normal to the z-direction to be “n”. Resolving the acceleration along the plane into n,z
components:
az = −a sin θ , an = a cos θ , θ = 20o
For rigid-body motion of the fluid in the n,z coordiantes:
dp = − ρ an dn − ρ ( g + az ) dz
dp = − ρ a cos θ dn − ρ ( g − a sin θ ) dz = 0 ← along free surface p = patm
Using trignometirc relationships this equation can be converted into x,y coordinates.
dn = dx cos θ + dy sin θ
dz = dy cos θ − dx sin θ
− ρ a cos θ dn − ρ ( g − a sin θ ) dz = 0
− ρ a cos θ [ dx cos θ + dy sin θ ] − ρ ( g − a sin θ ) [ dy cos θ − dx sin θ ] = 0
(
)
− ρ a cos 2 θ dx − ( ρ a cos θ sin θ ) dy −  ρ ( g − a sin θ ) cos θ  dy −  ρ ( g − a sinθ )( − sin θ )  dx = 0
 − ρ a cos2 θ + ρ ( g − a sin θ ) sin θ  dx +  − ρ a cos θ sin θ − ρ ( g − a sin θ ) cos θ  dy = 0




 − ρ a cos2 θ + sin2 θ + ρ g sin θ  dx + [ − ρ g cos θ ] dy = 0


[ − ρ a + ρ g sin θ ] dx − [ ρ g cos θ ] dy = 0
(
)
[ −a + g sin θ ] dx − [ g cos θ ] dy = 0
Integration yields:
( −a + g sin θ ) x − ( g cos θ ) y = −C

a
sin θ 
y = −
+
 x +C
 g cos θ cos θ 
The constant of integration can be determined by noting that the container is ½-full:
∀water =  ydx
0


sin θ 
a
=  ydx  −
+
 x + C  dx
 g cos θ cos θ 

0

 2
a
= −
+ tan θ  + C
 g cos θ
 2
 a

+
− tan θ 
 g cos θ
2


m
1 2


10.0 cm )( 6.0 cm ) / 2
(
o  (10.0 cm )
s

+
− tan 20
C=
2
  9.81 m  cos 20o

(10.0 cm )


2
s 


= 1.723 cm
C=
∀water
Solving for the requested length:


a
y = −
+ tan θ  x + C
 g cos θ



−1
o
+
h=
tan
20
 (10 cm ) + 1.723 cm = 4.277 cm
 ( 9.81) cos 20o



h = 4.28 cm
Problem 2.157
The U-tube manometer in the figure below is used to measure the acceleration of the cart
on which it sits. Develop an expression for the acceleration of the cart in terms of the liquid
height h, the liquid density ρ , the local acceleration of gravity g, and the length .
ℓ
h
p
g
a
Solution 2.157
patm
Writing Newton’s second law in the horizontal direction
(x-direction) for the bottom leg of the manometer gives
 Fx = ma x ,
h
hℓ
p A − pr A = ρ Aa,
hr
or
a=
p − pr
x
ρ
pℓ
Applying the manometer rule to the two legs of the manometer gives
p = patm + ρ gh
and
pr = patm + ρ gh
Subtracting gives
p − pr = ρ g ( h − hr ) = ρ gh
so
a=
ρ gh
ρ
or
h
a = g 
 
A
pr
Problem 2.158
A tank has a height of 5.0 cm and a square cross section measuring 5.0 cm on a side. The
tank is one-third full of water and is rotated in a horizontal plane with the bottom of the
tank 100 cm from the center of rotation and two opposite sides parallel to the ground.
What is the maximum rotational speed that the tank of water can be rotated with no water
coming out of the tank?
Solution 2.158
dp = − ρ gdz + ρω 2 r dr
Since dp = 0 along the free surface, the free surface is identified
by the equation
z
0 = − ρ gdz + ρω r dr
g–
h
2
(r1, –
or
b
)
2
b
r1 + h = 100 cm
0 = − gdz + ω 2 r dr
Integrating gives
z
r
b
2
r1
0 = − g  dz + ω 2  rdr ,
−
b ω

0 = −g  z +  +
r 2 − r12 ,
2 2

2
(
)
or
b ω2 2 2
z=− +
r − r1 .
2 2g
(
)
Recognizing that the volume of water in the rotating tank must equal
b2 h
gives
6
r
r +h
r1 + h
 b ω2 2 2 
b2 h 1
r − r1  dr ,
=  zb dr = b   − +
6
2
2
g



r
r
(
)
1
r +h
 br ω 2  r3 2   1
b2 h
= b − +
 − r1 r  
6
 2 2 g  3
 
r1
,
dr
 bh ω 2  ( r + h )3 r3

b2 h
 1
= b − +
− 1 − r12 h   ,

6
3
3
 2 2g 



3
2bh ω 2  ( r1 + h ) r13 2 

=
− − r1 h  ,

3
2g 
3
3


or
ω=
4bhg
 ( r + h )3 r3

− 1 − r12 h 
3 1


3
3


.
The numerical values give
ω=
cm 

4 ( 5 cm )( 5 cm )  981 2 
s 

 (100 cm )3 ( 95 cm )3

2
3
−
− ( 95 cm ) ( 5 cm ) 


3
3


rad   rev   60 s 

=  3.68


 or
s   2π rad   min 

ω = 35.1rpm
DISCUSSION Note the that when r = r1 + h,
)
(
b ω2
b ω2
z=− +
2r1h + h2 .
( r1 + h )2 − r12 = − +
2 2g
2 2g
(
)
The numerical values give
2
rad 

 3.68

5
s  
2
z = cm + 
2 ( 95 cm )( 5 cm ) + ( 5 cm ) 


cm 
2

2  981 2 
s 

= 4.23cm
or the assumption indicated in the above figures that the water does not reach the
uppermost side of the tank is correct.
Problem 2.159
An open 1-m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is
rotated about its vertical axis the center of the fluid surface is depressed. At what angular
velocity will the bottom of the tank first be exposed? No water is spilled from the tank.
Solution 2.159
fluid surface
Equation for surfaces of constant pressure:
z=
ω 2r2
2g
+ constant
ω 2r2
The volume of fluid in rotating tank is given by
R
2πω 2 3
πω 2 R 4
r
dr
V f =  2π rh dr =
=
2 g 0
4g
0
Since the initial volume, Vi = π R 2 hi , must equal the final volume,
V f = Vi
So that
πω 2 R 4
4g
x
y
R
hi ~ initial depth
2g
R
dr h
r
hi
For free surface with h = 0 at r = 0 ,
h=
z
= π R 2 hi
or
m

4  9.81 2  ( 0.7 m )
4 ghi
rad
s 

ω=
=
= 10.5
2
2
s
R
( 0.5 m )
Problem 2.160
The U-tube in the figure below rotates at 2.0 rev/sec. Find the
absolute pressures at points C and B if the atmospheric
pressure is 14.696 psia. Recall that 70°F water evaporates at an
absolute pressure of 0.363 psia . Determine the absolute
pressures at points C and B if the U-tube rotates at 2.0 rev/s.
ω
A
C
B
1′
2.5′
70°F water
Solution 2.160
patm
Applying the manometer rule to one of the legs and using
the data in Table A.6,
pB = patm +ρ gh
lb 

 62.3 3  (1in.)
ft 
pB = 14.696 psia + 
,

in.3 
1728 3 
ft 

h
C
B
R
pB = 14.732 psia
Section 2.6.2 gives:
∂p
= ρ rω 2.
∂r
pB
R
Integrating from r = 0 to r = R gives:  dp = ρω 2  r dr or
pC
pB − pC =
ρω 2 R 2
0
2
.
Therefore,
2
lbm  
rev 

2
62.3 3   2.0
2.5 ft )
(


2 2
ρω R
sec 
ft  

pc = pB −
= 14.732 psia –
2
2
 144 in.2   rev   32.2 ft ⋅ lbm 
2
 ft 2   2π rad   lb ⋅ sec 2 


pc = 8.10 psia
Check for phase change: pc > 0.363psia ➔ no evaporation ➔ above answer is correct.
DISCUSSION Note that if pc were calculated to be less than, some of the water would
vaporize and pc would be 0.363psia.
Problem 2.161
A child riding in a car holds a string attached to a floating, helium-filled balloon. As the car
decelerates to a stop, the balloon tilts backwards. As the car makes a right-hand turn, the
balloon tilts to the right. On the other hand, the child tends to be forced forward as the car
decelerates and to the left as the car makes a right-hand turn. Explain these observed effects
on the balloon and child.
Solution 2.161
FB-W = buoyant force – weight
A floating balloon attached to a string will align
p1
g
itself so that the string it normal to lines of constant
p2
p3
pressure. Thus, if the car is not accelerating, the
lines of p = constant are horizontal (gravity acts
T = tension in
constant
θ
string
vertically down), and the balloon floats “straight
pressure lines
up” (i.e., θ = 0). If forced to the side (θ ≠ 0 ), the
Fig. (1) No acceleration, θ = 0 for equilibrium.
balloon will return to the vertical (θ = 0 equilibrium
position in which the two forces T and FB -W line up.
Consider what happens when the car decelerates with an
amount ay < 0 .
As show by the equation,
slope =
car
deceleration
z
y
car
motion
p1
ay
dz
,
=−
dy
g + az
the lines of constant pressure are not horizontal,
ay
ay
dz
=−
=−
>0
dy
g + az
g
p2
p3
dz
dy
since az = 0 and ay < 0 .
Again, the balloon’s equilibrium position is with the string
normal to p = const. lines. That is, the balloon tilts back as
the car stops.
Fig. (2) Balloon aligned so that
string is normal to p = constant lines
g
ay = V /R
2
Fig. (3) Left turn; balloon tills to right
V2
When the car turns, ay =
(the centrifugal acceleration), the lines of p = const. are as
R
shown, and the balloon tilts to the outside of the curve.
Problem 2.162
A closed, 0.4-m-diameter cylindrical tank is completely filled with oil ( SG = 0.9 ) and rotates
about its vertical longitudinal axis with an angular velocity of 40 rad / s. Determine the
difference in pressure just under the vessel cover between a point on the circumference and
a point on the axis.
Solution 2.162
ω
B
A
z
y
x
0.2 m
Pressure in a rotating fluid varies in accordance with the equation,
p=
ρω 2 r 2
Since
2
− γ z + constant
=
pB − pA =
,
ρω 2
2
( rB2 − rA2 )
2
kg  rad 

(0.9) 103 3  40

s  
m 

(0.2m)2 − 0 
=


2
= 28.8 kPa
Problem 2.163
The largest liquid mirror telescope uses a 6-ft-diameter tank of mercury rotating at 7 rpm to
produce its parabolic-shaped mirror as shown in the figure below. Determine the difference
in elevation of the mercury, Δ h, between the edge and the center of the mirror.
Receiver
Light rays
6 ft
Δh
ω = 7 rpm
Mercury
Solution 2.163
For free surface of rotating liquid,
z=
ω r
2 2
2g
z
+ constant
Let z = 0 at r = 0 , and therefore, constant = 0 .
Thus, Δh = Δz for r = 3ft and with


rad
= 0.733
s
ω = ( 7 rpm )  2π
rad   1min 


rev   60 s 
It follows that
2
rad 
2

 0.733
 ( 3ft )
s 
Δh = 
= 0.0751ft
ft 

2  32.2 2 
s 

r
Problem 3.2
Air flows steadily along a streamline from point (1) to point (2) with negligible viscous
effects. The following conditions are measured: At point (1) z1 = 2 m and p1 = 0 kPa ; at
point (2) z2 = 10 m , p2 = 20 N/m 2 , and V2 = 0 . Determine the velocity at point (1).
Solution 3.2
(2)
z2 = 10 m
p2 = 20 N2
m
V2 = 0
air
ρ = 1.23
(1) z1 = 2 m
p1 = 0
V1 = ?
γ = ρg
kg
m3
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z 2
2
2
Thus, with p1 = 0 and V2 = 0 ,
1
ρV12 + γ z1 = p2 + γ z 2
2
or
1
kg 
N 
kg 
m
1.23 3  V12 = 20 2 + 1.23 3  9.81 2 (10 m − 2 m )

2

m 
m
m 
s
or
V12 =
2
2(20) N ⋅ m
m
m

+ 2 9.81 2  (8 m) = 189 2
1.23 kg
s 
s

Thus,
V1 = 13.7
m
s
 kg ⋅ m 
m

N ⋅ m  s2 
m2
=
= 2)
(Note:
kg
kg
s
Problem 3.3
Water flows steadily through the variable area horizontal pipe shown in the figure below.
The centerline velocity is given by V = 10(1 + x )ˆi ft , where x is in feet. Viscous effects are nes
∂p
glected. (a) Determine the pressure gradient,
(as a function of x), needed to produce
∂x
this flow. (b) If the pressure at section (1) is 50 psi, determine the pressure at (2) by
(i) integration of the pressure gradient obtained in (a), (ii) application of the Bernoulli’s
equation.
V(x)
Q
ℓ = 3 ft
(2)
(1)
x
Solution 3.3
(a) − γ sin θ − ∂ p = ρV ∂ V but θ = 0 and V = 10(1 + x ) ft
∂s
∂s
s
∂p
∂V
or ∂ p = − ρV ∂ V = − ρ (10(1 + x ))(10)
= − ρV
∂s
∂s
∂x
∂x
2
Thus, ∂ p = − 1.94 slugs 10 ft  (1 + x ) , with x in feet
∂x
ft 3 
= −194 (1 + x )
s
lb
ft 2
p2
(b) (i) ∂ p = − 194 (1 + x ) so that
∂x

p1 =50psi
x2 =3
dp = −194  (1 + x ) dx
x1 =0

32  lb  1ft 2 
or p2 = 50 psi − 194  3 +  2 
 = 50 − 10.1 = 39.9 psi

 ft  144 in.2 
2


(ii) p1 + 1 ρV12 + γ z1 = p2 + 1 ρV 22 + γ z 2
2
2
or with z1 = z2
ft
s
1
where
p 2 = p1 + ρ V12 − V 22
ft
2
V2 = 10(1 + 3) = 40
s
1
slugs 
ft 2  1ft 2 
Thus, p2 = 50 psi + 1.94 2  (102 − 402 ) 2 
 = 39.9 psi
2
ft 
s  144 in.2 
(
)
V1 = 10(1 + 0) = 10
Problem 3.4
What pressure gradient along the streamline, d p , is required to accelerate water in a
ds
horizontal pipe at a rate of 30 m / s 2 ?
Solution 3.4
∂p
∂ V where θ = 0 and ∂ V
m
= − γ sin θ − ρV
V
= as = 0 2
∂s
∂s
∂s
s
Thus,
∂p
kg  m 
 N  1
= − ρ a s = − 999 3  30 2  = − 30, 000  2 
∂s
m  s 
m m
kPa
∂p
= −30.0
m
∂s
Problem 3.5
m
and the pressure gradient along the streamline is
s
At a given location the airspeed is 20
100
N
m3
. Estimate the airspeed at a point 0.5 m farther along the streamline.
Solution 3.5
If neglect gravity, ∂ p = − ρV ∂ V or ∂ V = − ∂ p 1
∂s
∂s
∂s
∂ s ρV
N
100 3
1
∂V
m
=−
= −4.07
or
kg   m 
∂s
s

1.23
20 

3
m  s 
Thus,
δV ≈
∂V
1
m
m
m

δ s =  − 4.07  (0.5 m ) = − 2.03 , so that V + δ V = 20 − 2.03
s
s
∂s
s
s

or V ≈ 18.0
m
s
Problem 3.6
What pressure gradient along the streamline,
dp
, is required to accelerate water upward in a
ds
ft
vertical pipe at a rate of 30 2 ? What is the answer if the flow is downward?
s
Solution 3.6
∂p
∂V
= −γ sin θ − ρV
∂s
∂s
where θ = 90 for up flow,
θ = −90 for down flow,
and V ∂ V = a s = 30 ft
∂s
s2
Thus, for up flow
∂p
lb
slugs  ft 
psi
 lb  1
= −62.4(1) 3 − 1.94 3 30 2  = −120.6  2  = −0.838
∂s
ft
ft
ft  s 
 ft  ft
and for down flow
∂p
lb
slugs  ft 
psi
 lb  1
= −62.4( −1) 3 − 1.94 3 30 2  = 4.20  3  = 0.0292
∂s
ft
ft
ft  s 
 ft  ft
Problem 3.7
The Bernoulli’s equation is valid for steady, inviscid, and incompressible flows with constant acceleration of gravity. Consider flow on a planet where the acceleration of gravity
varies with height so that g = g0 − cz , where g0 and c are constants. Integrate " F = ma "
along a streamline to obtain the equivalent of the Bernoulli’s equation for this flow.
Solution 3.7
z
g
S
(2)
δm
(1)
From  δ Fs = δ mas one obtains dp + 1 ρ d (V 2 ) + γ dz where γ = ρ g
2
(see Equation along a streamline: dp + 1 ρ d (V 2 ) + γ dz = 0 )
2
Thus,
1

dp + d  ρV 2  + ρ ( g 0 − cz )dz = 0 , or by integrating from (1) to (2):
2


2
2
2
1
2
 dp +  d  2 ρV  + ρ  ( g0 − cz )dz = 0
1
1
1
or
p2 − p1 +
(
)
(
)
1
1
ρ V 22 − V12 + ρ g 0 ( z 2 − z1 ) − ρ c z 22 − z12 = 0
2
2
Thus,
p+
1
1
ρV 2 + ρ g 0 z − ρ cz 2 = constant alo ng a streamline.
2
2
Problem 3.8
An incompressible fluid flows steadily past a circular cylinder as shown in the figure
below. The fluid velocity along the dividing streamline ( −∞ ≤ x ≤ − a ) is found to be
 a2 
V =V0 1 − 2  , where a is the radius of the cylinder and V0 is the upstream velocity.
 x 


(a) Determine the pressure gradient along this streamline. (b) If the upstream pressure is
p0 , integrate the pressure gradient to obtain the pressure p( x) for ( −∞ ≤ x ≤ −a ) . (c) Show
from the result of part (b) that the pressure at the stagnation point ( x = −a ) is p0 + ρV02 / 2 ,
as expected from the Bernoulli’s equation.
Dividing
streamline
x=0
V0
p0
x
Stagnation
point
a
Solution 3.8
2
∂V ∂V ∂x ∂V
 −2  2a V0
=
=
= V0 [ −a2 ]  3  =
(a) ∂ p = − γ sin θ − ρV ∂ V but θ = 0 and
∂s ∂x ∂s ∂x
∂s
∂s
x3
x 
Thus,
  a 2  1
∂p
∂V
= − ρV
= −2 ρ a 2V02 1 −    3
∂s
∂x
  x   x
p
x
p0
x =−∞
(b)  dp =
dp
 dx dx
Or
p − p0 = −2ρa2V02
x 
2
x
 a   dx
−3
2 2
2 −5
 1−  x   x3 = −2ρa V0  [x − a x ]dx
−∞ 
−∞

Thus,
 a 2 1  a  4 
p = p0 + ρV02   −    for ( −∞ ≤ x ≤ −a )
 x  2  x  
(c) For x =−a , from part (b):
1
1


p x =− a = p0 + ρV02 ( −1)2 − ( −1)4  = p0 + ρV02
2
2


Note: Bernoulli’s equation from point (1) where V1 = V0 , p1 = p0 and z1 = z0 to point
(2) where V2 = 0, z2 = z0 gives
p1 +
1
1
ρV12 + γ z1 = p2 + ρV 22 + γ z 2
2
2
or
1
p2 = p0 + ρV02
2
(1)
X1 = – ∞
(2)
X2 = – a
Problem 3.9
Consider a compressible liquid that has a constant bulk modulus. Integrate " F = ma " along
a streamline to obtain the equivalent of the Bernoulli’s equation for this flow. Assume
steady, inviscid flow.
Solution 3.9
From the equation  dp + 1 V 2 + gz = C
ρ
2
1
dρ
dp + ρ d (V 2 ) + γ dz = 0 where γ = ρ g and dp = EV
where EV = bulk modulus =
2
ρ
constant
1
(see the equation p + ρV 2 + γ z = constant along the streamline )
2
Thus, along a streamline:
EV
EV
dρ
ρ
+
1
ρ d (V 2 ) + ρ gdz = 0 or
2
dρ
1

+ d  V 2  + gdz = 0 which can be integrated between points (1) and (2) to give
ρ
2

2
EV 
ρ2 d ρ
ρ1
V2  1
 z2
+  d  V 2  +  gdz = 0
V1  2
ρ
 z1
or
1
1 1
−EV  −  + [V22 −V12 ] + g[ z2 − z1 ] = 0
 ρ2 ρ1  2
Hence:
gz −
EV
ρ
+
V2
= constant along the streamline
2
Problem 3.11
Air flows along a horizontal, curved streamline with a 20ft radius with a speed of 110ft / s .
Determine the pressure gradient normal to the streamline.
Solution 3.11
−γ
dz
dz ∂p ρV 2
= 0 since the streamline is horizontal.
, where
−
=
dn
dn ∂n
R
R
V
n
Thus,
2


slug  
ft 

−  0.00238 3   100 
2


∂p
slug
lb
lb
ρV
s
ft  
 = −1.19
=−
= 
= −1.19 2 2 1
∂n
R
20 ft
ft ⋅ s   slug ⋅ ft  
ft3


  s 2  
Problem 3.12
Water flows around the vertical two-dimensional bend with circular streamlines and con-
stant velocity as shown in the figure below. If the pressure is 40 kPa at point (1), determine
the pressures at points (2) and (3). Assume that the velocity profile is uniform as indicated.
g
4m
(3)
2m
1m
V = 10 m/s
(2)
(1)
Solution 3.12
dz ∂p ρV 2
−γ
− =
with dz = 1 and V = 10 m .
dn ∂n
R
s
dn
Thus, with R = 6 − n ,
dp
ρV 2
= −γ −
, or
dn
6−n
n
n
n
dp
ρV 2dn
 dndn = −  γ dn −  6 − n
n=0
n=0
n=0
So that since γ and V are constants
n
p − p1 = −γ n − ρV
2
dn
 6−n
n=0
Thus,
 6 
p = p1 − γ n − ρV 2 ln 

6−n
N
kg  m 
With p1 = 40 kPa and n2 = 1m : p2 = 40 kPa − 9.8 ×103 3 (1m) − 999 3 10 
s
m
or p2 = 12.0 kPa
m 
2
6
ln  

5
and with p1 = 40 kPa and
N
kg  m 
n3 = 2 m : p3 = 40 kPa − 9.80 ×103 3 (2 m) − 999 3 10 
s
m
or p3 = −20.1 kPa
m 
2
6
ln  

4
Problem 3.13
Water flows around the vertical two-dimensional bend with circular streamlines as is shown
in the figure below. The pressure at point (1) is measured to be p1 = 25 psi and the velocity
across section a − a is as indicated in the table below. Calculate and plot the pressure across
section a − a of the channel [ p = p( z ) for 0 ≤ z ≤ 2 ft] .
z (ft)
V (ft/s)
0
0.2
0.4
0.6
0.8
1.0
0
8.0
14.3
20.0
19.5
15.6
1.2
1.4
1.6
1.8
2.0
8.3
6.2
3.7
2.0
0
g
20 ft
P1 = 25 psi
a
(1)
2 ft
V = V (z)
z
a
Solution 3.13
−γ
dz
dz ∂p ρV 2
= 1, R = 22 – n, and V = V ( n ) as given in the table with z = n .
with
−
=
dn
dn ∂n
R
Thus,
ρV 2
dp
= −γ −
dn
22 − n
or
p1
n =2
n =2
n
(22 − n )
n
 dp = −  γ dn − 
p
ρV 2
dn
or
n=2
V2
p1 − p = −γ (2 − n) − ρ 
dn
(22
−
n
)
n
Hence with γ = 62.4
lb
ft 3
, ρ = 1.94
slugs
ft 2
lb 
in.2 
lb
, and p1 = 25 2 144 2  = 3600 2 this gives
in. 
ft 
ft
2
V2
p = 3600 + 62.4(2 − n ) + 1.94 
dn , where p
22 − n
n
lb
ft 2
, n
(1)
ft
For 0 ≤ n ≤ 2 use the data in the table (V = V ( n ) , where n = z )
and integrate numerically to determine p = p( n ) .
z (ft)
V (ft/s)
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
0
8.0
14.3
20.0
19.5
15.6
8.3
6.2
3.7
2.0
0
n, (ft)
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
Value of integral
13.33
13.04
11.8
8.98
5.32
2.37
0.879
0.361
0.107
0.02
0
p, (lb/ft2)
3751
3738
3723
3705
3685
3667
3652
3638
3625
3613
3600
3800
3750
p (lb/ft2)
3700
3650
3600
0
0.5
1
n (ft)
1.5
2
Problem 3.14
Water in a container and air in a tornado flow in horizontal circular streamlines of radius r
and speed V as shown in the figure below. Determine the radial pressure gradient, dp / dr ,
ft
needed for the following situations: (a) The fluid is water with r = 3 in. and V = 0.8 .
s
(b) The fluid is air with r = 300 ft and V = 200 mph .
y
V
r
x
Solution 3.14
For curved streamlines,
dp ρV 2
dz
− =
+ γ , or with dz = 0 (horizontal streamlines), R = r, and d = − d
dn
R
dn
dn
dr
dn
this becomes
dp ρV 2
=
dr
r
With r =
3
ft
slugs
ft , V = 0.8 , and water  ρ = 1.94 3  ,
12
s
ft 

2
slugs 
ft 
1.94 3 0.8 
slugs
lb
dp
s
ft 
=
= 4.97 2 2 = 4.97 3
3
dr
ft ⋅ s
ft
ft
12
ft 

 88 s 
ft
With r = 300 ft, V = 200 mph 
 = 293 ,
s
 60 mph 


slug 

and air  ρ = 0.00238 3  ,
ft 

2
slugs 
ft 
0.00238 3 293 
dp
slugs
lb
s
ft 
=
= 0.681 2 2 = 0.681 3
dr
300 ft
ft ⋅ s
ft
Problem 3.15
Air flows smoothly over the hood of your car and up past the windshield. However, a bug
in the air does not follow the same path; it becomes splattered against the windshield. Explain why this is so.
Solution 3.15
An air particle flowing along streamline (1)–(2) is immersed in a pressure field produced by
all of the surrounding air particles. Gravity and pressure effects precisely balance centrifugal acceleration effects.
(2)
air particle
V
(1)
That is −γ
dz ∂p ρV 2
, where γ and ρ are the specific weight and density of the air.
−
=
dn ∂n
R
R
V
air particle follows
streamline past windshield
A bug is more dense than air, ρbug > ρ , but it “feels” the same pressure field, which is not
sufficient to make it turn as sharply as the air does. Hence, Rbug > R and the bug hits the
windshield.
air
bug
Problem 3.17
At a given point on a horizontal streamline in flowing air, the static pressure is −2.0 psi
(i.e., a vacuum) and the velocity is 150 ft/s. Determine the pressure at a stagnation point on
that streamline.
Solution 3.17
(1)
(2)
ft
s
p1 = –2.0 psi
V1 = 150
V2 = 0
z1 = z2
p1 +
1
1
ρV12 + γ z1 = p2 + ρV 22 + γ z 2
2
2
where z1 = z2 and V2 = 0
Thus,
2
1
lb  
in.2  1 
slugs  
ft 
2 
p2 = p1 + ρV1 =  −2.0 2   144 2  +  0.00238 3  150 
2
s
in.  
ft  2 
ft  



slug ⋅ ft  1lb 
= −2.88 2 + 26.8 2 2 

ft
ft ⋅ s  slug ⋅ ft 


 s2 
lb
= −261 2 = −1.81psi
ft
lb
Problem 3.19
When an airplane is flying 200 mph at 5000ft altitude in a standard atmosphere, the air
velocity at a certain point on the wing is 273 mph relative to the airplane. (a) What suction
pressure is developed on the wing at that point? (b) What is the pressure at the leading edge
(a stagnation point) of the wing?
Solution 3.19
1
(a) p + ρV 2 + z = constant
2
(3)
V3 = 273 mph
(2)
(1)
V2 = 0
V1 = 200 mph
ρ = 2.048 × 10−3
slug
ft3
Thus, with z1 ≈ z2 ≈ z3
1
1
p1 + ρV12 = p3 + ρV32 , but p1 = 0 so that
2
2
1
p3 = ρ [V12 −V32 ]
2
ft 

 88 s 
ft
where V1 = 200 mph 
 = 293
s
 60 mph 


ft 

 88 s 
ft
and V3 = 273 mph 
 = 400
s
 60 mph 


or
1
slugs 
ft 2
lb
p3 =  2.05 ×10 −3 3   2932 − 4002  3 = −76.0 2 (gage)


2
ft 
s
ft
(b) Also,
2
1
1
slugs 
ft 
lb
p2 = ρV12 =  2.05 ×10 −3 3   293  = 88.0 3 (gage)
2
2
s
ft  
ft
Problem 3.20
Air flows over the airfoil shown in the figure below. Sensors give the pressures shown at
points A, B, C , and D . Find the air velocities just above points A, B, C , and D . The air
density is 0.0020 slug / ft3 .
V0 = 100 mph
A
B
C
p0 = 13.20 psia
D
pA = 13.35 psia
pB = 12.5 psia
pC = 12.8 psia
pD = 13.0 psia
Solution 3.20
Assume inviscid, constant air density. Apply Bernoulli’s equation between point 0 and
point i (i = A, B, C , D ) ,
p0
ρ
+
V0 2
p V2
+ gz0 = i + i + gzi
2
ρ
2
Where z0 ≅ zi . Then
Vi =
2( p0 − pi )
ρ
+V02
The known values give
2(13.2 − pi )
Vi =
lb 
in2 
144


in2 
ft 2 
2
slug   lb ⋅ sec 
ft hr 

 0.002 3   slug ⋅ ft  1.47 sec mi 
ft  



+ (100 mph ) = 66,600(13.2 − pi ) + 10,000 mph
2
2
For point A,
VA = 66,600(13.2 − 13.35) + 10,000 mph
VA = 3.2 mph
For point B ,
VB = 66,600(13.2 − 12.5) + 10,000 mph
VB = 238 mph
For point C,
VC = 66,600(13.2 − 12.8) + 10,000 mph
VC = 191 mph
For point D,
VD = 66,600(13.2 − 13.0) + 10,000 mph
VD = 153 mph
Problem 3.21
Some animals have learned to take advantage of the Bernoulli’s effect without having read
a fluid mechanics book. For example, a typical prairie dog burrow contains two entrances—a flat front door and a mounded back door as shown in the figure below. When the
wind blows with velocity V0 across the front door, the average velocity across the back
door is greater than V0 because of the mound. Assume the air velocity across the back door
is 1.07V0 . For a wind velocity of 6 m/s, what pressure differences, p1 − p2 , are generated to
provide a fresh airflow within the burrow?
1.07 V0
V0
(2)
(1)
Q
Solution 3.21
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2
2
2
Thus, with negligible gravitational effect (i.e., z1 ≈ z2 )
2
2
1
1
kg   
 m    m  
2
2
p1 − p2 = ρ (V2 −V1 ) = 1.23 3   1.07  6   − 6  
2
2
 s    s  
m   
or p1 − p2 = 3.21
N
m2
Problem 3.23
2013 Indianapolis 500 champion Tony Kanaan holds his hand out of his IndyCar while
driving through still air with standard atmospheric conditions. (a) For safety, the pit lane
speed limit is 60 mph . At this speed, what is the maximum pressure on his hand? (b) Back
on the race track, what is the maximum pressure when he is driving his IndyCar at 225 mph ?
(c) On the straightaways, the IndyCar reaches speeds in excess of 235 mph . For this speed,
is your solution method for parts (a) and (b) reasonable? Explain.
Solution 3.23
(a)
p1 V12
p
V2
+
+ z1 = 2 + 2 + z2
γ 2g
γ
2g
with
z1 = z2

ft s 
ft
p1 = 0, V2 = 0
V1 = 60 mph  1.466
 = 88
mph 
s

Thus p2 =
γ
2g
V12 =
1
ρV12
2
2
p2 =
or
1
slug   ft 
lb
2.38 × 10 −3 3   88  = 9.22 2

2
ft   s 
ft

ft / s 
ft
(b) If V1 = 225 mph  1.466
 = 330 , then
mph 
s

2
p2 =
1
slug  
ft 
lb
2.38 × 10 −3 3   330  = 129.6 2

2
s
ft  
ft

ft / s 
ft
(c) Now V1 = 235 mph  1.466
 = 345
mph 
s

As related in section 3.8.1, flow can be modeled as incompressible if density change is less
than 3%. For air at standard conditions with isentropic flow, the velocity should be less
than approximately 335ft/s . Thus, using the (incompressible) Bernoulli’s equation may not
be reasonable at 345ft/s .
Problem 3.24
What is the minimum height for an oil ( SG = 0.75 ) manometer to measure airplane speeds
up to 30 m / s at altitudes up to 1500 m ? The manometer is connected to a Pitot-static tube
as shown in the figure below.
V = 30 m/s
h
SG = 0.75
Solution 3.24
Assume the air incompressible for these low speeds. Assuming frictionless flow, Bernoulli’s
equation gives from 1 to 2 (the stagnation point) for an observer on the airplane.
1
p2 = p1 + ρaV12
2
Interpolating from Table C.2 Properties of the U.S.
Standard Atmosphere (SI Units) gives
p1 = ( patm )1500 m = 8.469 ×10 4 Pa and
ρa = ( ρ )1500 m = 1.06
kg
m3
4
1 V1
2
.
pB
Then
pA
2
kg   m 

 1.06 3   30 
s 
m 
p2 = 8.469 × 10 4 Pa+ 
= 85200 Pa
 kg ⋅ m   N 
2


 N ⋅ s 2   m 2 ⋅ Pa 
oil
V1 = 30 m/s
Now pB ≈ p4 and p4 ≈ p1 assuming z4 − zB is small. Applying the manometer rule to the
manometer gives
p4 = p 2 − ρ oil gh
or
h=
p2 − p4
p2 − p1
=
ρoil g
Soil ρ H 2O g
The numerical values give
h=
(85200 − 84690 ) Pa 100
cm 

m 

kg  
m

0.75  1000 3   9.81 2 

m 
s 
h = 6.93 cm
Problem 3.25
A Pitot-static tube is used to measure the velocity of helium in a pipe. The temperature and
pressure are 40 F and 25 psia . A water manometer connected to the Pitot-static tube indicates a reading of 23in. Determine the helium velocity. Is it reasonable to consider the flow
as incompressible? Explain.
Solution 3.25
h
V
(1) (2)
p1 V12
p V2
+
+ z1 = 2 + 2 + z2
γ 2g
γ 2g
with z1 = z2 , V1 = V2 and V2 = 0
Thus, V1 = 2 g
( p2 − p1 )
γ
=
25
where ρ =
2( p2 − p1 )
ρ
lb 
in.2 
144


in.2 
ft 2 
p
slugs
=
= 5.80 × 10 −4 3
RT 
ft
4 ft ⋅ lb 
1.242 × 10 slugs ⋅ R  (460 + 40)°R


and since γ H2O
γ He
p2 − p1 = γ H 2O h = 62.4
lb  2.3 
lb
ft
11.96
=


ft 3  12 
ft 2
lb 

2 11.96 2 
ft  = 203 ft

Thus, V1 =
slugs
s
5.80 × 10 −4 3
ft
Note: Ma =
V
where c = kRT
c
1

2
ft ⋅ lb
ft
(460 + 40)°R  = 3210
Thus, c = 1.66(1.242 ×10 4 )
slug ⋅°R
s


ft
s = 0.063
or Ma =
ft
3210
s
203
0.3
Thus, the flow can be considered incompressible.
See also section 3.8.1 where V1 = 203
ft
ft
< 335 .
s
s
Problem 3.26
A Bourdon-type pressure gage is used to measure the pressure from a Pitot tube attached to
the leading edge of an airplane wing. The gage is calibrated to read in miles per hour at
standard sea level conditions (rather than psi). If the airspeed meter indicates 150 mph
when flying at an altitude of 10,000ft , what is the true airspeed?
Solution 3.26
(2)
V1
(1)
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 where p1 = 0 , V2 = 0 , z1 = z2
γ 2g
γ 2g
Thus, p2 =
1
ρV22
2
If the gage reads 150 mph then, since it is calibrated at sea level,
2
1
slugs  
mi  1hr  5280 ft  
lb
p2 =  0.00238 3  150


  = 57.6 3
2
hr  3600 s  mi  
ft  
ft
Hence, at 10,000 ft where ρ = 0.001756
57.6
1
slugs  ft 
=
0.001756

 V1 
ft 2 2 
ft3   s 
lb
or V1 = 256
slugs
ft 3
this pressure corresponds to
2
ft
ft  3600 s   1mi 
= 256 
 = 175 mph

s
s  hr   5280 ft 
Problem 3.28
A 40-mph wind blowing past your house speeds up as it flows up and over the roof. If ele-
vation effects are negligible, determine (a) the pressure at the point on the roof where the
speed is 60 mph if the pressure in the free stream blowing toward your house is 14.7 psia .
Would this effect tend to push the roof down against the house, or would it tend to lift the
roof? (b) Determine the pressure on a window facing the wind if the window is assumed to
be a stagnation point.
Solution 3.28
V2
(2)
V1
(3)
(1)
The Bernoulli’s equation gives
1
1
1
p1 + ρV12 = p2 + ρV22 = p3 + ρV32
2
2
2
a) Thus, from (1) to (2):
1
1
mi  1hr   5280 ft 
ft
p1 + ρV12 = p2 + ρV22 or with V1 = 40 

 = 58.7
2
2
hr  3600s   mi 
s
and V2 = 60
mi
ft
= 88.0
hr
s
2
1
slugs  
ft  
ft  
p2 = p1 +  0.00238 3   58.7  −  88.0  
2
s 
s  
ft  
2
or
p2 − p1 = −5.12
lb
ft 2
This negative pressure tends to lift the roof.
b) From (1) to (3): Since V3 = 0 ,
1
p3 = p1 + ρV12 or
2
2
1
1
slugs 
ft 
lb
p3 − p1 = ρV12 =  0.00238 3   58.7  = 4.10 2
2
2
s
ft  
ft
Problem 3.29
Pressurized eyes Our eyes need a certain amount of internal pressure in order to work
properly, with the normal range being between 10 and 20 mm of mercury. The pressure is
determined by a balance between the fluid entering and leaving the eye. If the pressure is
above the normal level, damage may occur to the optic nerve where it leaves the eye,
leading to a loss of the visual field termed glaucoma. Measurement of the pressure within
the eye can be done by several different noninvasive types of instruments, all of which
measure the slight deformation of the eyeball when a force is put on it. Some methods use a
physical probe that makes contact with the front of the eye, applies a known force, and
measures the deformation. One noncontact method uses a calibrated “puff” of air that is
blown against the eye. The stagnation pressure resulting from the air blowing against the
eyeball causes a slight deformation, the magnitude of which is correlated with the pressure
within the eyeball. (See Problem 3.29.)
Determine the air velocity needed to produce a stagnation pressure equal to 10 mm of
mercury.
Solution 3.29
N
1
ρV 2 = pstag = 10 mm of mercury=γ Hg h , where γ Hg = 133 ×103 3
2
m
 1m  
1
kg 
3 N 
Thus, 1.23 3 V 2 = 10 mm 
 133 ×10 3 
2
m 
m 
 1000 mm  
or V = 46.5
m
s
Problem 3.30
Water flows through a hole in the bottom of a large, open tank with a speed of 8
Determine the depth of water in the tank. Viscous effects are negligible.
Solution 3.30
p1 = 0
(1)
V1 = 0
z1 = h
h
(2)
V2 = 8 m
s
p2 = 0
z2 = 0
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2
2
2
Thus, with p1 = p2 = z2 = V1 = 0 ,
1
2
γ z1 = ρV22 , where γ = ρ g and z1 = h
so that ρ gh =
1
ρV22
2
2
 m
2
8 
V2
s 
or h =
= 
= 3.26 m
m
2g

2  9.81 2 
s 

.
Problem 3.32
The tank shown in the figure below contains air at atmospheric pressure above the water
surface. The velocity of the water flowing from the tank is 22.5ft / sec . Determine the water
level ( h ).
D = 12 in.
Air
p = 14.7 psia
h
V2 = 22.5 ft/sec
d2
Solution 3.32
Assume quasi-steady state and constant water density. Apply Bernoulli’s equation to a
streamline connecting the water surface (1) to the exit (2),
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2
ρ
2
ρ
2
Now p1 = p2 = patm and assume V1 ≈ 0 . This gives
g ( z1 − z2 ) =
V22
,
2
2
ft 

22.5
2


V2
sec 

,
h=
=
ft 
2g

2  32.2

sec 2 

or
h = 7.86 ft
Problem 3.33
Water flows from the faucet on the first floor of the building shown in the figure below with
a maximum velocity of 20ft/s . For steady inviscid flow, determine the maximum water
velocity from the basement faucet and from the faucet on the second floor (assume each
floor is 12ft tall).
4 ft
8 ft
4 ft
V = 20 ft/s
12 ft
4 ft
Solution 3.33
p V2
+
+ z = constant
γ 2g
Thus,
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 with p2 = p1 = 0 (free jet)
γ 2g
γ 2g
and V1 = 20
or
2
 ft 
 20 
V22
s

+ 4 ft =
+ ( −8 ft)
ft 
ft 


2  32.2 2 
2  32.2 2 
s 
s 


or V2 = 34.2
ft
s
ft
, z1 = 4 ft , z2 = −8ft
s
and
p V2
p1 V12
+
+ z1 = 3 + 3 + z3
γ 2g
γ 2g
with p3 = p1 = 0 (free jet)
and V1 = 20
ft
, z1 = 4 ft , z3 = 16 ft
s
2
 ft 
 20 
V32
s
or 
+ 4 ft =
+ 16 ft
ft 
ft 


2  32.2 2 
2  32.2 2 
s 
s 


or V3 = 202 − 2(32.2)(12) = −373
Impossible! No flow from second floor faucet.
Problem 3.34
Laboratories containing dangerous materials are often kept at a pressure slightly less than
ambient pressure so that contaminants can be filtered through an exhaust system rather
than leaked through cracks around doors, etc. If the pressure in such a room is 0.1in. of
water below that of the surrounding rooms, with what velocity will air enter the room
through an opening? Assume viscous effects are negligible.
Solution 3.34
V1 = 0
V2
(2)
(1)
If viscous effects are negligible,
1
1
p1 + ρV12 = p2 + ρV22 , where V1 = 0 and p1 − p2 = γ H2O h
2
2
or
lb 
lb
 0.1  
p1 − p2 = 
ft   62.4 3  = 0.520 2
 12  
ft 
ft
Thus,
1
 
lb   2
1
2  0.520 2  

 2( p − p )  2
ft   = 20.9 ft
V2 =  1 2  =  
s
ρ
 0.00238 slugs 


3 

ft 
Problem 3.36
Streams of water from two tanks impinge upon each other as shown in the figure below. If
viscous effects are negligible and point A is a stagnation point, determine the height h.
h
Free jets
A
p1 = 25 psi
Air
20 ft
8 ft
Solution 3.36
p2
γ
+
V22
p
V2
+ z2 = A + A + zA , where p2 = 0 , V2 = 0 , z2 = h + 20 ft , VA = 0 , and zA = 20 ft
2g
γ
2g
Thus,
h + 20 ft =
pA
γ
+ 20 ft
or
h=
pA
(1)
γ
Also,
p1 V12
p
V2
+
+ z1 = A + A + zA , where p1 = 25 psi , V1 = 0 , and z1 = 8ft
γ 2g
γ
2g
Thus,
pA
γ
=
p1
γ
+ z1 − zA which when combined with Eq. (1) gives
25
h=
p1
γ
+ z1 − zA =
lb 
in.2 
144


in.2 
ft 2 
+ 8ft − 20 ft = 45.7 ft
lb
62.4 3
ft
Problem 3.37
Several holes are punched into a tin can as shown in the figures below. Which of the figures
represents the variation of the water velocity as it leaves the holes? Justify your choice.
(a)
(b)
(c)
(b)
(c)
Solution 3.37
(1)
h
(2)
(a)
p V2
+
+ z = constant , so that with V1 = 0 , p1 = 0 , and z1 = h1 at the free surface, then
γ 2g
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 or with p2 = 0 (free jet) and z2 = h2
γ 2g
γ 2g
or
h1 =
V22
+ h2 so that V2 = 2 g ( h1 − h2 ) = 2 gh
2g
Thus,
V2
O
V2
or
h
O
h
Fig. (a) is correct distribution
Problem 3.38
Water flows from a pressurized tank, through a 6-in. -diameter pipe, exits from a 2-in. diameter nozzle, and rises 20ft above the nozzle as shown in the figure below. Determine
the pressure in the tank if the flow is steady, frictionless, and incompressible.
20 ft
2 in.
Air
2 ft
6 in.
Solution 3.38
(2)
20 ft
2 in.
Air (1)
2 ft
6 in.
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where V1 = 0 , V2 = 0 , z1 = 2 ft , z2 = 22 ft , and p2 = 0
γ 2g
γ
2g
Thus,
p1
γ
= z 2 − z1
or
lb 
lb

p1 = γ ( z2 − z1 ) =  62.4 3  (22 ft − 2 ft) = 1248 2
ft 
ft

Note: The diameter of the pipe or nozzle is not needed.
Problem 3.39
The piston shown in the figure below is forcing 70°F water out the exit at 12 ft/sec. The exit
pressure has been measured as pe = 40 psig . Determine the force on the piston for a piston
diameter Dp = 3.0 in .
F
V = 2 ft/sec
h = 8.5 ft
g
pe = 40 psig
Assume constant density and inviscid flow and apply Bernoulli's equation to a streamline
from a point just below the piston ( p ) to exit ( e),
pp Vp2
pe Ve2
+
+ gze =
+
+ gz p
ρ 2
ρ
2
or
ρ
pp = pe + Ve2 −Vp2 + ρ g ze − z p .
2
(
)
(
)
Table B.1 gives ρ = 1.936 lbm/ft3 so
2
slug  2

2 ft
1.936 3  12 − 2
ft 
sec 2
p p = 40 psig + 
 slug ⋅ ft  
in2 
2 
144

2
 
ft 2 
 lb ⋅ sec  
(
)
+
slug  
ft 

1.936 3   32.2
 ( −8.5 ft )
ft  
sec 2 

 slug ⋅ ft  
in2 
144


2
 
ft 2 
 lb ⋅ sec  
= ( 40 + 0.94 − 3.68 ) psig = 37.3psig
Assuming there is no friction between the piston and the cylinder wall and that the piston
weight is negligible,
F = pg
π Dp2
4
F = 264 lb.
lb  π ( 3.0 in )

=  37.3 2 
4
in 

2
Problem 3.41
Air flows steadily through a horizontal 4-in. -diameter pipe and exits into the atmosphere
through a 3-in. -diameter nozzle. The velocity at the nozzle exit is 150ft/s . Determine the
pressure in the pipe if viscous effects are negligible.
Solution 3.41
V2 = 150 ft/s
(1)
air
V1
(2)
p2 = 0
3 in.
4 in.
ρ = 0.00238
z1 = z2
slugs
ft3
From Bernoulli’s equation,
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2
2
2
Thus, with z1 = z2 , p2 = 0 , and V2 = 150
p1 =
(
1
ρ V22 −V12
2
ft
,
s
)
π 2 
2
2
 4 D2 
 D2 
ft 
ft
 3 in.  
=
A
V
But, AV
,
or
V
V
V
=
=
=
150  = 84.4

1 1
2 2

1 
2 
2 


π
s
s
 4 in.  
 D1 
 D12 
4

Thus,


2
2
1
slugs  
ft  
ft  
slug  1lb 
p1 =  0.00238 3  150  −  84.4   = 18.3


2
s 
s  
ft  
ft ⋅ s2  slug ⋅ ft 


 s2 
or
p1 = 18.3
lb
ft 2
= 0.127 psi
Problem 3.42
The figure below shows a tube for siphoning water from an aquarium. Determine the rate
at which the water leaves the aquarium for the conditions shown. Is there an advantage to
having the large-diameter section? The water flow is inviscid.
¼ in. I.D. tube
h = 5 ft
2
A = 6 ft
Solution 3.42
Denote the aquarium free surface by 1 and the tube outlet by 2. Assuming constant fluid
density, Bernoulli’s equation between point 1 and point 2 gives
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2
ρ
2
ρ
2
Now p1 = p2 = patm and V1 << V2 since A >> Atube . Then
ft 
ft

V2 = 2 ( gz1 − z 2 ) = 2 gh = 2  32.2 2  ( 5ft ) = 17.9
sec
sec 

The rate at which water is drained out of the tank Q is
π
2  17.9ft  
2
ft   60sec 
Q = A2V2 =
V2 = (0.25 in.) 
 


4
4
 sec   12 in.   min 
π d22
ft3
Q = 0.366
min
If the water were inviscid, the large diameter tube would not affect the discharge rate.
Since the water is somewhat viscous, the large diameter tube would reduce the viscous
resistance to fluid flow. It also reduces the inlet velocity and makes it less likely to pick up
small pebbles and/or fish.
Problem 3.43
For the pipe enlargement shown in the figure below, the pressures at sections (1) and (2) are
56.3 and 58.2 psi , respectively. Determine the weight flowrate (lb/s) of the gasoline in the
pipe.
Q
2.05 in.
Gasoline
3.71 in.
(1)
(2)
Solution 3.43
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
where z1 = z2 , and AV
1 1 = A2V2
2
D 
or V2 =  1  V1
 D2 
Thus,
2
 D1  2

 V1
2
p1 V1
p2  D2 
+
=
+
γ 2g γ
2g
or
1
 
ft
lb
lb  
in2   2
 2  32.2 
58.2 2 − 56.3 2  144 2  
2 

s
in.
in.  
ft  


2 g ( p2 − p1 )
V1 =
=

4
  D 4  




lb
2.05
in.

42.5 3 1 − 
γ 1 −  1   


ft   3.71in.  
  D2   



or
2
V1 = 21.4
ft
π  2.05  
ft 
ft3
=
=
ft
21.4
0.490
and Q = AV
1 1

 

s
4  12  
s
s
Thus,
lb 
ft3 
lb
γ Q = 42.5 3  0.490  = 20.8
s 
s
ft 
Problem 3.44
1
A fire hose nozzle has a diameter of 1 in. According to some fire codes, the nozzle must be
8
capable of delivering at least 250 gal/min. If the nozzle is attached to a 3-in. -diameter hose,
what pressure must be maintained just upstream of the nozzle to deliver this flowrate?
Solution 3.44
D1 = 3 in.
D2 = 1.125 in.
V1
(1)
(2)
V2
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , with z1 = z2 , p2 = 0
γ 2g
γ 2g
gal  
in3  1ft3   1min 
ft3

Q
=
250
231
=
0.557
and




3

min  
gal 
s

 1728 in   60 s 
Thus,
p1 =
γ 
V22 − V12 


2g
ft3
0.557
Q
ft
s
= 80.7
where V2 = =
2
A2 π  1.125  2
s
ft


4  12 
0.557
and V1=
ft 3
s
Q
ft
=
= 11.34
2
A1 π  3  2
s
  ft
4  12 
so that with
γ
g
=ρ
1
slugs 
ft 2
p1=  1.94 3  80.7 2 − 11.34 2  2
s
2
ft  
lb
= 6190 2 = 43.0 psi
ft
Problem 3.45
Water flowing from the 0.75-in. -diameter outlet shown in the figure below rises 2.8in.
above the outlet. Determine the flowrate.
2.8 in.
0.75 in.
Q
Solution 3.45
The flowrate is Q = AV
1 1 where from the Bernoulli’s equation
p1 V12
p V2
+
+ z1 = 2 + 2 + z2
γ 2g
γ 2g
Thus, with p1 = p2 = z1 = V2 = 0 we obtain
ft   2.8 
ft

V1 = 2 gz2 = 2  32.2 2  
ft  = 3.88
s
s   12 

So that
2
π  0.75  
ft 
ft 3
Q = AV
ft   3.88  = 0.0119
1 1=

4  12  
s
s
Problem 3.46
A fire hose has a nozzle outlet velocity of 30 mph . What is the maximum height the water
can reach?
Solution 3.46
The maximum height is when the water velocity is directed upward. Apply Bernoulli’s
equation between the nozzle outlet (0) and the maximum height (m). Assume inviscid flow.
p0
ρ
+
V02
p
V2
+ gz0 = m + m + gz m
2
ρ
2
Now p0 = pm = patm , set z0 = 0 , and Vm = 0 . This gives
2
2
ft   hr 
 mi  
 30
  5280
 

2
mi   3600 sec 
V
 hr  
zm = 0 =
ft 
2g

2  32.2

sec 2 

zm = 30.1ft
2
Problem 3.47
At what rate does oil (SG
0.85) flow from the tank shown in the figure below?
D = 14 cm
10.5 cm
13.5 cm
dT = 3.0 cm
Oil
dB = 3.0 cm
Solution 3.47
Apply Bernoulli’s equation between the oil surface (0) and the outlet (T ).
p0
ρ
+
V02
p
V2
+ gz0 = T + T + gzT
2
ρ
2
dT2 , assume V0 = 0 . Also p0 = pT = patm and set z0 = 0 . This gives
Since D 2
m
cm 
cm


VT = 2 gz0 = 2  9.81 2  (10.5cm ) 100
.
 = 144
m
s

s 

Now apply Bernoulli’s equation between the oil surface (0) and the outlet (B).
p0
ρ
+
V02
p
V2
+ gz0 = B + B + gzB .
2
ρ
2
Again note V0 = 0 and p0 = pB = patm and set zB = 0 .
Then
m
cm 
cm


VB = 2 gzB = 2  9.81 2  (13.5cm ) 100
.
 = 163
m
s

s 

Since AT = AB , the total flow rate is
Q = VT AT +VB AB = AT (VT +VB )
=
π
4
(3cm )2 (144 + 163)
cm
cm3
= Q = 2170
s
s
Problem 3.48
Find the water height hB in tank B shown in the figure below for steady-state conditions.
hA = 10.0 cm
dA = 0.01 m
Tank A
hB
dB = 0.02 m
Tank B
Solution 3.48
Assume constant fluid density and inviscid flow and apply Bernoulli’s equation to tank A
from the surface (1) to the outlet (2).
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2 .
ρ
2
ρ
2
Now p1 = p2 = patm and assume V1 ≈ 0 .
Setting z2 = 0 gives z1 = hA = 10 cm .
Then
m  10 
m

V2 = 2 ghA = 2  9.81 2  
m  = 1.4 .
s
s   100 

and
m π
m

2
.
Q = V2 A2 =  1.4  ( 0.01m ) = 1.1 × 10 −4
s 4
s

3
From the continuity, the flow rate leaving tank A (and entering tank B) must equal the rate
leaving tank B. So
3

−4 m 
1.1
10
×


s 
m
Q 
VB =
=
= 0.35
π
s
AB
( 0.02 m )2
4
Now apply Bernoulli’s equation to tank B from the surface (3) and the outlet (4).
V32
p4 V42
+
+ gz3 =
+
+ gz4 .
ρ
2
ρ
2
p3
Assume V3 ≈ 0 , set z4 = 0 so z3 = hB , and p3 = p4 = patm
Then
2
m

0.35 
2
2

V
V
cm 
s 

hB = 4 = B = 
= hB = 0.00624 m  100

m
2g 2g
m 


2  9.81 2 
s 

hB = 0.624 cm
Problem 3.49
The pressure and average velocity at point A in the pipe shown in the figure below are
16.0 psia and 4.0 ft / sec , respectively. Find the height h and the pressure and average
velocity at point B. Fluid fills the 1-in. -diameter discharge pipe.
60 °F water
h
D
A
5 ft
d = 1 in.-diameter plastic pipe
B
patm
Solution 3.49
Assume incompressible, inviscid flow. Appling Bernoulli’s equation to a streamline from
the free water surface (0) to point A gives
p0
ρ
+
V02
p
V2
+ gz0 = A + A + gzA .
2
ρ
2
Now V0 ≈ 0 and z is measured positive upward. So
h = z 0 − zA =
p0 − pA VA2
.
+
ρg
2g
The numerical values give
=

 














+ 









=
Since the fluid fills the pipe,
VB = VA = 4.0
ft
sec
Now apply Bernoulli’s equation between points A and B.
pA
ρ
+
VA2
p
V2
+ gzA = B + B + gzB .
2
ρ
2
Since VA = VB ,
=
+ρ
(
pB = 3.47 psig
−
)=


+

(




)



Problem 3.50
Water (assumed inviscid and incompressible) flows steadily in the vertical variable-area
pipe shown in the figure below. Determine the flowrate if the pressure in each of the gages
reads 50kPa.
2m
10 m
p = 50 kPa
1m
Q
Solution 3.50
2m
(1)
V1
10 m
p = 50 kPa
(2)
V2
1m
Q
From the Bernoulli’s equation,
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2 , where p1 = p2 = 50 kPa
2
2
Thus,
1
ρ (V22 −V12 ) = γ ( z1 − z2 )
2
(1)
Also, AV
1 1 = A2V2 , or
π 2 
2
2
 4 D2 
 D2 
 1m 
A2
1
V1 = V2 = 
V2 = 
V2 = 
V2 = V2



π
A1
4
 2m 
 D1 
 D12 
4

Hence, Eq. (1) becomes
1  2 1 2
ρ V2 − V2  = ρ g ( z1 − z2 )
2 
16

or
15 2
m

V2 = 2 g ( z1 − z2 ) = 2  9.81  (10 m)
16
s 

or
V2 = 14.5
m
s
Thus,
Q = A2V2 =
π
m
m3

(1m)2 14.5  = 11.4
4
s 
s

Problem 3.51
Air is drawn into a wind tunnel used for testing automobiles as shown in the figure below
(a) Determine the manometer reading, h, when the velocity in the test section is 60 mph .
Note that there is a 1-in. column of oil on the water in the manometer. (b) Determine the
difference between the stagnation pressure on the front of the automobile and the pressure
in the test section.
Wind tunnel
60 mph
Fan
Open
h
1 in.
Oil (SG = 0.9)
Water
Solution 3.51
Wind tunnel
60 mph
(2)
(1)
(3)
Fan
Open
h
Water
1 in.
Oil (SG = 0.9)
p1 V12
p2 V22
(a)
+
+z =
+
+ z , where z1 = z2 , p1 = 0 , and V1 ≈ 0
γ 2g 1 γ 2g 2
Thus, with
V2 = 60 mph = 88
ft
,
s
V22
=−
or
γ
2g
p2
2
1
1
slugs  ft 
lb
p2 = − ρV22 = −  0.00238 3   88  = −9.22 2
2
2
ft   s 
ft
lb 
lb

 1 
But p2 + γ H 2O h − γ oil  ft  = 0 where γ oil = 0.9γ H 2O = 0.9  62.4 3  = 56.2 3
 12 
ft 
ft

Thus,
−9.22
lb
ft
2
+ 62.4
lb
ft
3
( h ft) − 56.2
lb  1 
 ft  = 0 , or h = 0.223ft
ft 3  12 
(b)
p2
γ
+ z2 +
V2
V22 p3
=
+ z3 + 3 , where z2 = z3 , and V3 = 0
2g γ
2g
Thus,
p2 V22 p3
+
=
or
γ 2g γ
2
1
1
slugs  ft 
lb
p3 − p2 = ρV22 =  0.00238 3   88  = 9.22 2
2
2
ft   s 
ft
Problem 3.52
The figure below shows a duct for testing a centrifugal fan. Air is drawn from the
atmosphere ( patm = 14.7 psia, Tatm = 70°F). The inlet box is 4 ft × 2 ft . At section 1, the duct
is 2.5ft square. At section 2, the duct is circular and has a diameter of
. A water
manometer in the inlet box measures a static pressure of −2.0 in. of water. Calculate the
volume flow rate of air into the fan and the average fluid velocity at both sections 1 and 2.
Assume constant density.
2 in. water
Throttling
plug
Inlet box
(1)
(2)
Fan
Solution 3.52
The pressure in the inlet box
is
lb   2 

 62.4 3   12 ft 
 = 14.63psia .
ft  
pi = patm − ( ρ g )water h = 14.7 psia − 
2

in 
144 2 
ft 

Now apply Bernoulli’s equation from the atmosphere to the inlet .
p0
ρa
+
V02
p V2
+ gz0 = i + i + gzi .
2
ρa 2
Now p0 = patm , assume V0 ≈ 0 and z0 ≈ zi , and use Table B.3 Physical Properties of Air at
Standard Atmospheric Pressure to get ρa = 0.0023
Vi =
2 ( p0 − pi )
ρa
2 (14.7 − 14.63)
=
slug
ft 3
lb 
in2 
144


in2 
ft 2 
slug   lb ⋅ s2 

0.00233



ft3   slug ⋅ ft 

Then
ft 

Qi = Vi Ai =  93.0  ( 4 ft × 2 ft )
s

. Then
ft3
Qi = 744
s
= 93.0
ft
.
s
The continuity equation gives
ft3
Q
s
V1 = i =
A1 ( 2.5 ft )2
744
V1 = 119
and
ft3
744
Q
s
V2 = i =
A2 π 3.0 ft 2
(
)
4
V2 = 105
ft
s
ft
s
Problem 3.53
Natural gas (methane) flows from a 3-in. -diameter gas main, through a 1-in. -diameter
ft3
pipe, and into the burner of a furnace at a rate of 100 . Determine the pressure in the gas
hr
main if the pressure in the 1-in. pipe is to be 6 in. of water greater than atmospheric
pressure. Neglect viscous effects.
Solution 3.53
3 in.
1 in.
(1)
(2)
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2 ,
2
2
where z1 = z2 , and V1 =
(1)
Q
Q
, V2 =
A1
A2
with
Q = 100
ft 3 1hr
ft3
×
= 0.0278
and
hr 3600 s
s
=




=

Thus,
ft3
s =0.566 ft and
V1 =
2
s
π3 
ft


4  12 
0.0278
ft 3
0.0278
s =5.10 ft
V2 =
2
s
π1 
ft


4  12 
Hence, since ρ = 1.29×10−3
slugs
ft 3
from Table 1.7, Eq. (1) becomes
1
p1 = p2 + ρ (V22 −V12 )
2
2
2

1
ft  
ft  
−3 slugs  
= 31.2 2 + 1.29 ×10
  5.10 s  −  0.566 s  
2
 
 
ft
ft3  
lb
= (31.2 + 0.0165)
or
p1 = 31.217
lb
ft 2
lb
ft 2
Problem 3.54
Air flows radially outward between the two parallel circular plates shown in the figure
=
below. The pressure at the outer radius
is atmospheric. Find the pressure at the
inner radius =
if the air density is constant, the air flow is inviscid, the volume flow
rate is 4.0 L / s , and the plate spacing is =
.
po = patm = 101.3 kPa
T = 20°C
Vi
h
Vo
Ri
Ro
g
Solution 3.54
Apply Bernoulli’s equation from radius
to radius R0.
Vi 2
p0 V02
+
+ gzi =
+
+ gz0
ρ
2
ρ
2
pi
Since zi = z0,
pi = p0 +
(
ρ V02 −Vi2
2
).
The volume flow rate Q gives
2
L   −3 m 3  
cm 

4.0
10
100






s  
L  
m 

Q
m
Vi =
=
= 102
2π Ri h
2π ( 0.5 cm )( 0.125 cm )
s
and
2
L   −3 m3  
cm 

 100
 4.0  10

s 
L  
m 

Q
m
=
= 10.2
V0 =
2π R0 h
2π ( 5.0 cm )( 0.125 cm )
s
Table B.4 or the ideal gas law gives ρ = 1.204
pi = 101.3kPa +
pi = 95.1kPa
kg
m3
so
2
2
kg  
m 
m    N ⋅ s2 


1.204 3  10.2  − 102   
s  
s    kg ⋅ m 

m  

N / m2 
2 103


kPa 

Problem 3.55
Air flows radially inward between the two parallel circular plates shown in the figure below.
The pressure at the outer radius R0 = 5.0 cm is atmospheric. Find the pressure at the inner
radius Ri = 0.5 cm if the air density is constant, the air flow is inviscid, the volume flow rate
is 4.0 L / s , and the plate spacing is h = 0.125 cm .
po = patm = 101.3 kPa
T = 20°C
Vi
h
Vo
Ri
Ro
g
Solution 3.55
Apply Bernoulli’s equation from radius
pi
ρ
+
to radius
.
Vi 2
p V2
+ gzi = 0 + 0 + gz0
2
ρ
2
Since zi = z0 a 2 + b 2 ,
pi = p0 +
(
ρ V02 −Vi2
2
),
The volume flow rate Q gives
2
L   −3 m 3  
cm 

  100
 4.0   10

s 
L  
m 

Q
m
Vi =
=
= 102
2π Ri h
2π ( 0.5 cm )( 0.125 cm )
s
and
2
L   −3 m3  
cm 

4.0
10
100




s  
L  
m 

Q
m
=
= 10.2
V0 =
2π R0 h
2π ( 5.0 cm )( 0.125 cm )
s
Table B.4 or the ideal gas law gives ρ = 1.204
pi = 101.3kPa +
kg
m3
so
2
2
kg  
m 
m    N ⋅ s2 


1.204 3  10.2  − 102   
s  
s    kg ⋅ m 

m  

N / m2 
2 103


kPa 

pi = 95.1kPa
Therefore,
pi = 95.1 kPa as in Problem 3.54 and the pressure difference (p0 − pi) is the same regardless
of whether the flow is radially inward or radially outward (for inviscid flow).
Problem 3.56
Find the water mass flow rate at the nozzle outlet O shown in the figure below, and calculate
the maximum height to which the water stream will rise. The water density is 1.9slugs ft3 ,
and the flow is inviscid.
h1 = 21 ft
d = 1.0 in.
O
h2 = 6 ft
A = 20 ft2
Solution 3.56
Denote the tank free surface by the subscript 1 and the nozzle outlet by the subscript 2.
Applying Bernoulli’s equation from 1 to 2 gives
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2
ρ
2
ρ
2
where p1 = p2 and V1 ≈ 0 . This gives
ft 
ft

V2 = 2 g ( z1 − z2 ) = 2 g ( h1 − h2 ) = 2  32.2 2  ( 21ft − 6 ft ) = 31.1
.
sec
sec 

The water mass flow rate is
m = ρ A2V2 =
m = 0.322
ρπ d 22V2
4
ft 
2
 slug 
 1.9 3  π (1.0 in )  31.1 sec 


ft 
=
2

in 
4  144 2 

ft 

slug
sec
Denote the nozzle outlet by the subscript 2 and the maximum height location by 3.
Applying Bernoulli’s equation from 2 to 3 gives
p V2
p2 V22
+
+ gz2 = 3 + 3 + gz3 .
2
2
ρ
ρ
Now p 2 = p3 and V3 = 0 at the maximum height location. Therefore,
2
ft 

31.1
2


V
sec 
z3 − z2 = 2 = 
ft 
2g

2  32.2

sec 2 

z3 − z2 = 15.0 ft = same height as tank free surface level.
Should you be surprised by this answer?
Problem 3.57
Water (assumed frictionless and incompressible) flows steadily from a large tank and exits
through a vertical, constant diameter pipe as shown in the figure below. The air in the tank
kN
is pressurized to 50 2 . Determine (a) the height h, to which the water rises, (b) the water
m
velocity in the pipe, and (c) the pressure in the horizontal part of the pipe.
Pipe
exit
50 kN/m2
h
Air
4m
2m
Water
Solution 3.57
(2)
Pipe
exit
(3)
50 kN/m2
h
Air
(0)
4m
Water
2m
(1)
a)
p0 V02
p V2
kN
+
+ z0 = 2 + 2 + z2 , where p0 = 50 2 , V0 = 0 , z0 = 2 m , p2 = 0 , and V2 = 0
γ 2g
γ 2g
m
Thus,
p
z2 = 0 + z0 or
γ
N
50 × 103 2
m + 2 m = 5.10 m + 2 m = 7.12 m
h=
3 N
9.8 × 10
m3
b) Since AV
1 1 = A3V3 and A1 = A3 , then V1 = V3 where
p3 V32
p V2
+
+ z3 = 2 + 2 + z2 with p3 = p2 = 0 , V2 = 0 .
γ 2g
γ 2g
Thus,
1
1
2
 

m
m
+ z3 = z2 or V3 = 2 g ( z2 − z3 ) 2 = 2  9.81 2  (7.12 − 4) m  = 7.82
s
2g
V32
[
Hence, V1 = V3 = 7.82
c)
]
 
s 

m
s
p V2
p1 V12
+
+ z1 = 3 + 3 + z3 , where V1 = V3 and p3 = 0 .
γ 2g
γ 2g
Thus,
p1 = γ ( z3 − z1 ) = 9.80 × 103
N
m
3
(4 − 0) m = 39.2 × 103
N
m2
= 39.2 kPa
Problem 3.58
ft
from the
s
large tank shown in the figure below. Determine the depth, H , of the layer of light liquid
lb
(specific weight = 50 3 ) that covers the water in the tank.
ft
Water (assumed inviscid and incompressible) flows steadily with a speed of 10
50 lb/ft3
10 ft/s
H
5 ft
Water
4 ft
Solution 3.58
From Bernoulli’s equation,
p1 V12
p2 V22
+
+z =
+
+ z , where p1 = γ 0 H , V1 = 0 , p2 = 0 , z1 = 4 ft , and z2 = 5 ft
γ 2g 1 γ 2g 2
Thus,
ft
γ0
V2
H + z1 = 2 + z2 so that with V2 = 10 ,
s
2g
γ
2
lb 
 ft 

10 
 50 ft3 
 s  + 5 ft

 H + 4 ft =
lb
ft 

 62.4

2  32.2 2 
3

ft 
s 

Therefore,
H = 3.19 ft
Problem 3.59
Water flows through the pipe contraction shown in the figure below. For the given 0.2-m
difference in the manometer level, determine the flowrate as a function of the diameter of
the small pipe, D.
0.2 m
0.1 m
Q
D
Solution 3.59
0.2 m
X
0.1 m
Q
(2)
D
(1)
p1 V12
p V2
+
+ z1 = 2 + 2 + z2
γ 2g
γ 2g
with z1 = z2 , and rearrangement,
V12 p2 V22 p1
=
+
− . With
2g γ
2g γ
p1
γ
p2
= x and
γ
+
V22
= 0.2 + x,
2g
V12
= 0.2 + x − x or V1 = 2 g ( 0.2 )
2g
1
 
2
m
m
V1 =  2  9.81 2  ( 0.2 )  = 1.98
s
s 
 

Thus,
Q = A1V1 =
π
4
( 0.1m )2 1.98

m
m3
0.0156
for any D
=

s 
s
Problem 3.60
Carbon tetrachloride flows in a pipe of variable diameter with negligible viscous effects. At
point A in the pipe, the pressure and velocity are 20 psi and 30 ft/s, respectively. At location
B the pressure and velocity are 23 psi and
by how much?
. Which point is at the higher elevation and
Solution 3.60
pA VA2
p V2
lb
+
+ zA = B + B + zB , with γ = 99.5 3
γ 2g
γ
2g
ft
or
2
lb 
in.2 
(20 − 23) 2 144 2  (302 −142 ) ft
in. 
ft 
p − p V 2 −VB2
s2
zB − zA = A B + A
=
+
lb
ft 
γ
2g

99.5 3
2  32.2 2 
ft
s 

or
zB − zA = 6.59 ft , B is above A
Problem 3.61
Water flows from a 20-mm-diameter pipe with a flowrate Q as shown in the figure below.
Plot the diameter of the water stream, d, as a function of distance below the faucet, h, for
values of 0 ≤ h ≤ 1m and 0 ≤ Q ≤ 0.004m3 /s. Discuss the validity of the one-dimensional
assumption used to calculate d = d ( h ) , noting, in particular, the conditions of small h and
small Q .
Q
20 mm
h
d
Solution 3.61
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = p2 = 0 , z1 = h and z2 = 0
γ 2g
γ 2g
Thus,
1
Q
Q
m
2
=
= 3183Q , if Q
V2 = (V1 + 2 gh ) 2 , where V1 =
A1 π (0.020 m)2
s
m3
s
4
Also, V2 =
Q
4Q
=
A2 π d 2
So that
4Q
πd2
(
2
= (3183Q )
1
+ 2(9.81)h 2
)
or
d=
2 Q
, where d
1
m, h
m, and Q
π 1.013 × 107 Q 2 + 19.62 h  4
Plot d = d ( h) for values of Q from Q = 0 to Q = 0.004
Note: Only the results for Q = 0.001
m3
(1)
s
m3
.
s
m3
m3
and Q = 0.004
are listed.
s
s
For Q = 0.00100 m3/s
h, m
d, m
For Q = 0.00400 m3/s
h, m
d, m
0.00000
0.25000
0.50000
0.75000
1.00000
1.25000
1.50000
1.75000
2.00000
2.25000
2.50000
2.75000
3.00000
3.25000
3.50000
0.00000
0.25000
0.50000
0.75000
1.00000
1.25000
1.50000
1.75000
2.00000
2.25000
2.50000
2.75000
3.00000
3.25000
3.50000
0.02000
0.01812
0.01689
0.01598
0.01528
0.01471
0.01423
0.01382
0.01346
0.01315
0.01287
0.01261
0.01238
0.01217
0.01198
0.02000
0.01985
0.01971
0.01957
0.01944
0.01931
0.01918
0.01906
0.01895
0.01883
0.01872
0.01861
0.01851
0.01841
0.01831
d, m
0
1
0
0.01
0.02
m3
Q = 0.001 s
h, m 2
m3
Q = 0.004 s
3
4
Note: From Eq. (1), as Q → 0 (with h ≠ 0 ) d →
but d = 0.02 m with h = 0 for any Q .
2 Q
1
19.62h 4
π[
]
→0
Problem 3.62
A liquid stream directed vertically upward leaves a nozzle with a steady velocity V0 and
cross-sectional area A 0. Find the velocity V and cross-sectional area A as a function of the
vertical position z .
Solution 3.62
Apply the continuity equation to a control volume enclosing the stream. Assuming constant
fluid density
A0V0 = AV
or
A 
V = V0  0 
 A
(1)
Appling Bernoulli’s equation from y = 0 to y gives
p0
ρ
+
V02
p V2
+ gy0 = +
+ gy .
2
ρ 2
Since p0 = p and y0 = 0 ,
V = V02 − 2 gy .
Using Eq. (1),
A=
A0V0
A0V0
=
=A
2
V
V0 − 2 gy
Problem 3.63
Water leaves a pump at 200 kPa and a velocity of 12 m / s . It then enters a diffuser to
increase its pressure to 250 kPa . What must be the ratio of the outlet area to the inlet area
of the diffuser?
Solution 3.63
Assume constant density and apply Bernoulli’s equation between 1 and 2 to get
p1 V12
p2 V22
+
+ gz1 =
+
+ gz2
ρ
2
ρ
2
Assume z1 ≈ z2 . This gives
1
 2 ( p1 − p2 )
2
V2 = 
+ V12 
ρ


(1)
The continuity equation gives
AV
1 1 = A2V2 so
A 
V2 = V1  1 
 A2 
(2)
Equating Eqs. (1) and (2) gives
1
2

N


2 ( 200 − 250 ) kPa  1000 2

 
A1  2 ( p1 − p2 )
⋅
m
kPa

 + 1

=
+ 1 =
2
2


A2  ρV1

kg   m 



1000
12






m3   s 
1
2
A1
= 0.553
A2
or
A2
= 1.81
A1
Problem 3.64
Water flows upward through a variable area pipe with a constant flowrate, Q , as shown in
the figure below. If viscous effects are negligible, determine the diameter, D( z ), in terms of
D1 if the pressure is to remain constant throughout the pipe. That is, p( z ) = p1 .
D(z)
z
D1
(1)
Q
Solution 3.64
p1 V12
p V2
+
+z = +
+ z , or with p = p1, and z1 = 0
γ 2g
γ 2g 1
V12
V2
+z =
, or V12 −V 2 = 2 gz
2g
2g
But V1 =
Q
4Q
Q
4Q
and
=
V
=
=
A π D2
A1 π D12
Thus,
2
 4Q   4Q 2
−
= 2 gz

2
  π D2 
D
π
 1 
or
1
1
π 2 gz
D1
D
8Q2
−
4
=
4
Thus,
D=
D1
1
4 4
gD1 z
2 
 π
1 −
8Q

2

Problem 3.65
The circular stream of water from a faucet is observed to taper from a diameter of 20 mm
to 10 mm in a distance of 50 cm. Determine the flowrate.
Solution 3.65
Q
(1)
D1 = 0.020 m
0.50 m
(2) D2 = 0.010 m
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = p2 = 0 , z2 = 0 , z1 = 0.50 m
γ 2g
γ 2g
Q
Q
and V1 =
, V2 =
A1
A2
Thus,
1
2



Q 
Q 
A 2 gz1
2 gz1


= 2
  + 2 gz1 = 
 or Q = 

2


 A1 
 A2 
 A2 
 1 − 1  
1− 

  A22 A12  
 A1 
or since
2
2
2
A2  D2 
=
 we obtain
A1  D1 
1
Q = A2
 
2
m
 2  9.81 2  (0.50 m) 
2 gz1
π
s 

= (0.010 m)2  
4


4
4
 0.010 
 D2 


−
1


1− 



 0.020 
D
 1
= 2.54 ×10 −4
m3
s
Problem 3.66
Water is siphoned from the tank shown in the figure below. The water barometer indicates
a reading of 30.2 ft. Determine the maximum value of h allowed without cavitation
occurring. Note that the pressure of the vapor in the closed end of the barometer equals the
vapor pressure.
Closed end
3-in.
diameter
30.2 ft
6 ft
h
5-in. diameter
Solution 3.66
Closed end
(0)
3-in.
diameter
(2)
30.2 ft
6 ft
(1)
h
5-in. diameter
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = 0 , V1 = 0 , p2 = pvapor , z1 = 0, z2 = 6 ft
γ 2g
γ 2g
Thus,
pvapor V22
0=
+
+ 6 ft
2g
γ
But p0 + 30.2 ft γ = p1 or since p0 = pvapor ,
pvapor
γ
= −30.2 ft
Hence,
0 = −30.2 ft +
V22
+ 6 ft
2g
or
V22
= 24.2 ft
2g
 

ft 
or V22 = 2  32.2 2  (24.2 ft) 
s 
 

Thus,
V2 = 39.5
ft
s
2
 3 in.  
A
D2
ft 
Since V3 A3 = V2 A2 , V3 = 2 V2 = 22 V2 = 
  39.5 
A3
s
D3
 5 in.  
or
V3 = 14.2
ft
s
However,
p3 V32
p1 V12
+
+z =
+
+ z or V3 = 2 gh
γ 2g 1 γ 2g 3
Thus,
14.2
ft
ft 

= 2  32.2 2  h ft
s
s 

or h = 3.13 ft
Problem 3.67
Water is siphoned from a tank as shown in the figure below. Determine the flowrate and
the pressure at point A, a stagnation point.
0.04-m diameter
3m
A
Solution 3.67
0.04 m diameter
(1)
3m
(2)
A
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = p2 = 0 , z1 = 3m , z2 = 0 , and V1 = 0
γ 2g
γ 2g
Thus,
m
m

V2 = 2 gz1 = 2  9.81 2  (3 m) = 7.67
s
s 

or
Q = A2V2 =
π
m
m3

(0.4 m )2  7.67  = 9.64 ×10 −3
4
s 
s

Also,
p1 V12
pA VA2
+
+z =
+
+ z where VA = 0 and z A = 0
γ 2g 1 γ 2g A
Thus,
pA
γ
= z1 or pA = γ z1 = 9.80
kN
m3
(3 m) = 29.4 kPa
Problem 3.68
A 50-mm -diameter plastic tube is used to siphon water from the large tank shown in the
figure below. If the pressure on the outside of the tube is more than 35 kPa greater than the
pressure within the tube, the tube will collapse and siphon will stop. If viscous effects are
negligible, determine the minimum value of h allowed without the siphon stopping.
2m
4m
h
Solution 3.68
(2)
2m
(1)
(3)
4m
h
At any location within the tube V = V3 so that with V1 = 0 , p1 = 0 and z1 = 0
V2
p1 V12
p V2
p
+
+ z1 = +
+ z gives = − z − 3 .
γ 2g
γ 2g
γ
2g
Thus, the lowest pressure occurs at the point of maximum z . That is, p2 = −35kPa and
=
so that
N
m
V32
p V2
p V2
m 2 = −2 m −
or V3 = 5.55
but 1 + 1 + z1 = 3 + 3 + z3
−
N
m
s
γ 2g
γ 2g

9.80 ×103 3
2  9.81 2 
m
s 

35 × 103
2
m

 5.55 
s 
where z3 = − ( 4 − h ) and p3 = 0 . Thus, 0 = 
− ( 4 − h ) or h = 2.43m
m

2  9.81 2 
s 

Problem 3.69
Water is siphoned from the tank shown in the figure below. Determine the flowrate from
the tank and the pressure at points (1), (2), and (3) if viscous effects are negligible.
2-in.-diameter
hose
2 ft
(3)
3 ft
8 ft
(2)
(1)
Solution 3.69
2-in.-diameter
hose
2 ft
(0)
(3)
3 ft
8 ft
(5)
(2)
(1)
V02
p V2
+ z0 = i + i + zi , for i = 1,...,5
γ 2g
γ 2g
For i = 5 and p0 = 0, V0 = 0 , p5 = 0 this becomes
p0
+
z0 =
V52
+ z5 or
2g
=
−
=






(1)
=
Thus,
Q = A5V5 =
π2
2
ft 
ft 3
 
ft
0.303
13.9
=

 

4  12  
s
s
From Eq. (1) with i
From Eq. (1) with i
lb 
lb

and V1 = 0, p1 = γ ( z0 − z1 ) =  62.4 3  (8ft) = 499 2
ft 
ft

2, z0 =
p2
γ
+
V22
+ z2
2g
where A2V2 = A5V5
Since A2 = A5 it follows that V2 = V5 or
V22 V52
=
= z0 − z5
2g 2g
Thus,
p2
γ
= z0 − z2 −
V2
V22
= z0 − z2 − 5 = z0 − z2 − ( z0 − z5 ) = z5 − z2
2g
2g
or
lb 
lb

p2 = γ ( z5 − z2 ) =  62.4 3  (5 ft) = 312 2
ft 
ft

p3 V32
From Eq. (1) with i = 3, z0 =
+
+z
γ 2g 3
where A3V3 = A5V5
Since A3 = A5 it follows that V3 = V5 or
V32 V52
=
= z0 − z5
2g 2g
Thus,
p3
γ
= z0 − z3 −
V32
V2
= z0 − z3 − 5 = z0 − z3 − ( z0 − z5 ) = z5 − z3
2g
2g
or
lb 
lb

p3 = γ ( z5 − z3 ) =  62.4 3  ( −3ft) = −187 2
ft 
ft

Problem 3.70
Water is siphoned from the tank shown in the figure below. A 1-in. -diameter nozzle is
placed at the end of the tube. Determine the flowrate from the tank and the pressure at
points (1), (2), and (3) if viscous effects are negligible.
2-in.-diameter
hose
2 ft
(3)
3 ft
8 ft
(2)
(1)
Solution 3.70
2-in.-diameter
hose
2 ft
(0)
(3)
3 ft
8 ft
(5)
(2)
p0
γ
+
(1)
V02
p V2
+ z0 = i + i + zi , for i = 1,...,5
2g
γ 2g
(1)
For i = 5 and p0 = 0, V0 = 0 , p5 = 0 this becomes
z0 =
ft 
ft
V52

+ z5 or V5 = 2 g ( z0 − z5 ) = 2  32.2 2  (3 ft) = 13.9
s
2g
s 

Thus,
Q = A5V5 =
π1
2
ft 
ft 3
 
ft
0.0758
13.9
=

 

4  12  
s
s
lb 
lb

From Eq. (1) with i = 1 and V1 = 0, p1 = γ ( z0 − z1 ) =  62.4 3  (8ft) = 499 2
ft 
ft

From Eq. (1) with i = 2, z0 =
V22
+
+z
γ 2g 2
p2
where A2V2 = A5V5
2
2
 D2 
1
2
Since A2 = 
 A5 =   A5 = 4 A5 it follows that V2 = V5 or
4
1
 D5 
2
V22
1 1 
1 V52 1
V
=
=
= ( z0 − z5 )
 5
2 g 2 g  4  16 2 g 16
Thus,
p2
γ
= z0 − z2 −
V22
1
1
= z0 − z2 − ( z0 − z5 ) = 8ft − (3ft) = 7.81ft
2g
16
16
or
lb 
lb

p2 =  62.4 3  (7.81ft) = 488 2
ft 
ft

From Eq. (1) with i = 3, z0 =
p3 V32
+
+z
γ 2g 3
where A3V3 = A5V5
or since A3 = A2 then V3 = V2 and
V32 V22 1
=
= ( z0 − z5 )
2 g 2 g 16
Thus,
p3
γ
= z0 − z3 −
V32
1
1
3
= z0 − z3 − ( z0 − z5 ) = − (3ft) = − ft
2g
16
16
16
or
lb  3 
lb

p3 =  62.4 3   − ft  = −11.7 2
ft   16 
ft

Problem 3.71
Water exits a pipe as a free jet and flows to a height h above the exit plane as shown in the
figure below. The flow is steady, incompressible, and frictionless. (a) Determine the height h
(b) Determine the velocity and pressure at section (1).
h
V = 16 ft/s
6-in. diameter
8 ft
(1)
4-in. diameter
Solution 3.71
( ) From the Bernoulli’s equation,
p2
γ
+
p V2
V22
+ z2 = 3 + 3 + z3 , where p2 = p3 = 0, and V3 = 0.
2g
γ 2g
Thus,
V22
= z3 − z2 = h
2g
or
2
 ft 
 16 
2
V2
s
= 
= 3.98ft
h=
ft 
2g

2  32.2 
s

(b) Also, AV
1 1 = A2V2
or
π
(6 in.)2
A2
ft
 ft 
4
V1 = V2 =
16  = 36.0

π
A1
s
(4 in.)2  s 
4
From the Bernoulli’s equation,
p1 V12
p2 V22
+
+z =
+
+z ,
γ 2g 1 γ 2g 2
or since γ = ρ g,
1
p1 = p2 + ρ (V2 −V1 ) + γ (z2 − z1 ) where p2 = 0
2
Thus,
2
2
1
slugs   ft  
ft  
lb
p1 = 1.94 3  16  −  36.0   + 62.4 3 (8ft)
2
s  
ft   s  
ft
lb
 slugs ⋅ ft  1
499
= −1009 
+

ft 2
 s2
 ft 2
= −510
lb
ft 2
Problem 3.72
Water flows steadily from a large, closed tank as shown in the figure below. The deflection
in the mercury manometer is 1 in. and viscous effects are negligible. (a) Determine the volume flowrate. (b) Determine the air pressure in the space above the surface of the water in
the tank.
Air
1-ft diameter
3-in. diameter
8 ft
1 in.
Mercury
Solution 3.72
(a) From the Bernoulli’s equation,
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where V2 = 0 and z1 = z2
γ 2g
γ 2g
(1)
Also, for the manometer,
p2 + γ H2O h = p1 + γ H2O ( h − lin.) + γ Hg (lin.)
or
p2 − p1 = (γ Hg − γ H2O )(lin.) = γ H2O (SGHg − 1)(lin.)
lb 

 1 
=  62.4 3  (13.56 − 1)  ft  = 65.3 ft
ft 
 12 

Thus, from Eq. (1),
lb
65.3 2
V12 p2 − p1
ft = 1.046 ft
=
=
2g
γ
62.4
lb
ft 2
(4)
Air
1-ft diameter
3-in. diameter
8 ft
(2)
(1)
(3)
so that
ft 
ft

V1 = (2)  32.2 2  (1.046 ft) = 8.21
s
s 

h
1 in.
Mercury
Hence,
π
ft 
ft3
Q = AV
1 1 = (1ft)  8.21  = 6.45
4
s
s

2
(b) From the Bernoulli’s equation,
p V2
p4 V42
Q
+
+ z4 = 3 + 3 + z3 , where V4 = 0 , p3 = 0 and V3 =
γ 2g
γ 2g
A3
Thus,
ft 3
s = 131 ft
V3 =
2
s
π3 
ft


4  12 
6.45
Hence, from Eq. (2),
1
V32
+ z4 =
+ z3 , or p4 = ρV32 + γ ( z3 − z4 )
2
γ
2g
p4
Hence,
2
1
slug 
ft 
lb
lb
p4 = 1.94 3  131  + 62.4 3 ( −8ft)=16150 2 = 112 psi
2
s
ft  
ft
ft
(2)
Problem 3.73
ft3
from a 3-in. pipe in which the pressure and tempers
ature are 20 psi (gage) and 120°F into a 1.5-in. pipe. If viscous effects are neglected and incompressible conditions are assumed, determine the pressure in the smaller pipe.
Carbon dioxide flows at a rate of 1.5
Solution 3.73
(1)
(2)
D2 = 1.5 in.
D1 = 3 in.
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
where z1 = z2 and
ft 3
Q
ft
s
V1 =
=
= 30.6
2
A1 π  3  2
s
  ft
4  12 
1.5
ft 3
Q
ft
s
V2 =
=
= 122
2
A2 π  1.5  2
s
ft


4  12 
1.5
Thus,
(
(20 + 14.7)
)
lb 
in.2 
144


in.2 
ft 2 
1
p
slug
p2 = p1 + ρ V12 −V22 where ρ =
=
= 7.62 ×10 −3 3
2
RT 
ft ⋅ lb 
ft
1130 slug ⋅ °R  (460 + 120)°R


or
2
2

lb 
in.2  1 
ft  
ft  
−3 slugs  
p2 = 20 2 144 2  +  7.62 ×10
  30.6 s  − 122 s  
in. 
ft  2 
ft3  
 
 
= 2880
lb
ft
2
− 53.1
lb
ft
2
or
p2 = 19.63 psi gage
= 2827
lb
ft 2
Problem 3.74
Oil of specific gravity 0.83 flows in the pipe shown in the figure below. If viscous effects are
neglected, what is the flowrate?
4 in.
Water
SG = 0.83
3 in.
Q
4 in.
Solution 3.74
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where z1 = z2 and V1 = 0
γ 2g
γ 2g
Thus,
V22 p1 − p2
=
2g
γ
(1)
but
4 in.=
p1 = p3 + γ = p4 + γ
(3)
Water
γ
and
(4)
γm
ℓ
SG = 0.83
3 in.
p2 = γ ( + h ) − γ m h + p4
h
(1)
(2)
Q
Thus,
p1 − p2 = (γ m − γ )h
(2)
Combine Eq. (1) and (2) to obtain


lb
62.4 3

 4
p −p 
γ

ft 



ft
− 1  ft 
V2 = 2 g  1 2  = 2 g  m − 1h = 2  32.2 2  
s   0.83  62.4 lb    12 

 γ 
 γ


 

ft3  


4 in.
or
V2 = 2.10
ft
s
Thus,
π 4
2
ft 
ft3
 
Q = A2V2 =  ft   2.10  = 0.183
4  12  
s
s
Problem 3.75
Water flows steadily through the variable area pipe shown in the figure below with negligible viscous effects. Determine the manometer reading, H , if the flowrate is 0.4 m 3 / s and
the density of the manometer fluid is 500 kg / m3 .
Density = 500 kg/m3
H
Area = 0.05 m2
Area = 0.07 m2
Solution 3.75
Density = 500 kg/m3
H
h
(1)
(2)
Area = 0.05 m2
Area = 0.07 m2
From the Bernoulli’s equation,
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where z1 = z2 .
γ 2g
γ 2g
Thus,
p2 − p1 =
γ
1
V12 −V22 ) = ρ (V12 −V22 )
(
2g
2
(1)
m3
m3
0.4
Q
s = 8 m and V = Q =
s = 5.71 m
=
But, Q = AV
2
1 1 = A2V2 so that V1 =
A1 0.05 m 2
A2 0.07 m 2
s
s
0.4
Hence, from Eq. (1)
 kg ⋅ m 
2
2
 2 

1
kg   m  
m
N
p2 − p1 =  999 3   8  −  5.71   = 15.7 ×103  s 2  = 15.7 ×103 2
2
s  
m   s  
m
m
(2)
For the manometer,
p1 − γ H 2O h − γ man H = p2 − γ H 2
( h − H ) so that
(
)
(
)
p2 − p1 = γ H2O ( h + H ) − γ H2O h − γ man H = γ H2O − γ man H = g ρH2O − ρman H
Hence, from Eqs. (2) and (3):
15.7 × 103
m
kg
kg 
= 9.81 2  999 3 − 500 3  H or H = 3.21m
m
s 
m
m 
N
2
(3)
Problem 3.76
The specific gravity of the manometer fluid shown in the figure below is 1.07. Determine
the volume flowrate, Q , if the flow is inviscid and incompressible and the flowing fluid is
(a) water, (b) gasoline, or (c) air at standard conditions.
Q
0.09-m
diameter
0.05 m
20 mm
10 mm
Solution 3.76
p1 V12
p2 V22
+
+z =
+
+z ,
γ 2g 1 γ 2g 2
where z1 = z2
and V2 = 0
(1)
Q
γ
ℓ
(2)
0.05 m
0.09-m
diameter
Thus,
V1 = 2 g
10 mm
( p2 − p1 )
(1)
γ
But
p1 + γ + γ m h = p2 + γ ( + h )
or
p2 − p1 = (γ m − γ )h
γm
so that Eq. (1) becomes

N  

1.07  9.8 ×103 3  

(γ − γ )
m

m  − 1 (0.02 m)

V1 = 2 g m
h = 2  9.81 2  
γ
γ

s 





Thus,
Q = AV
1 1=
π
4
D12V1 =
 10.49 ×103 
(0.09 m)2 2(9.81) 
− 1 (0.02)


4
γ


π
or
Q = 3.99 ×10−3
10 ⋅ 49
γ
−1
m3
where γ
s
kN
m3
20 mm = h
For the given fluids this gives:
fluid
γ
(a)
water
9.80
1.06 × 10−3
(b)
gasoline
6.67
3.02 × 10−3
(c)
air
12 × 10−3
0.118
=========
Problem 3.77
Water flows steadily with negligible viscous effects through the pipe shown in the figure below. It is known that the 4-in. diameter section of thin-walled tubing will collapse if the
pressure within it becomes less than 10 psi below atmospheric pressure. Determine the maximum value that h can have without causing collapse of the tubing.
4 ft
4-in.-diameter thin-walled tubing
h
6 in.
Solution 3.77
V2 p
V2
+ z1 + 1 = 2 + z2 + 2 ,
2g γ
2g
γ
p1
where
p1 = 0, V1 = 0,
z2 = 0 and p2 = −10
lb 
in.2 
lb
= −1440 2
144

2
2 

in 
ft 
ft
Thus, with z1 = 4 ft
(1)
lb
V22
ft 2 +
4 ft =
lb
ft 

62.4 3
2  32.2 2 
ft
s 

−1440
4 ft
h
4-in.-diameter thin-walled tubing
(2)
6 in.
ft
or V2 = 41.7
s
(3)
Also,
V32
V12 p3
+ z1 +
=
+ z3 +
,
γ
2g γ
2g
p1
where
2
2
 4 in.  
AV  D 
ft 
ft
p3 = 0, z3 = −h, and V3 = 2 2 =  2  V2 = 
  41.7  = 18.5
A3
s
s
 6 in.  
 D3 
Thus,
2
ft 

 18.5 
s
4 ft = − h + 
ft 

2  32.2 2 
s 

or
h = 1.31ft
Problem 3.78
Helium flows through a 0.30-m- diameter horizontal pipe with a temperature of 20 C and a
kg
pressure of 200 kPa (abs) at a rate of 0.30 . If the pipe reduces to 0.25-m-diameter, des
termine the pressure difference between these two sections. Assume incompressible,
inviscid flow.
Solution 3.78
(1)
Q
(2)
D1 = 0.3 m
D2 = 0.25 m
p1 = 200 kPa (abs)
T1 = 20°C
p1 V12
p2 V22
+
+z =
+
+ z , where z1 = z2
γ 2g 1 γ 2g 2
Thus,
N
200 ×103 2
1
p
m
p1 − p2 = ρ V22 −V12 where ρ = 1 =
2
RT1 
N⋅m 
 2077 kg ⋅ K  (273 + 20) K


(
)
or ρ = 0.329
kg
m3
Also,
m = ρ AV
1 1 = 0.30
kg
s
so that
kg
0.3
m
m
s
V1 =
=
= 12.9
kg  π
s
ρ A1 
2
 0.329 3  (0.3 m)
m 4

(1)
and
AV
1 1 = A2V2 or
2
2
D 
 0.3m  
m
m
V2 =  1  V1 = 
 12.9  = 18.6
s 
s
 0.25 m  
 D2 
Thus, from Eq. (1):
1
kg 
m2
p1 − p2 =  0.329 3  (18.62 − 12.92 ) 2 = 29.5 Pa
2
m 
s
Problem 3.79
ft3
Water is pumped from a lake through an 8-in. pipe at a rate of 10
. If viscous effects are
s
negligible, what is the pressure in the suction pipe (the pipe between the lake and the pump)
above the lake?
at an elevation
Solution 3.79
(2)
Q
(1)
p1 V12
p2 V22
+
+z =
+
+z ,
γ 2g 1 γ 2g 2
where p1 = 0, V1 = 0, z1 = 0, z2 = 6.0 ft
and
 ft3 
4 10


s 
Q
4Q
ft

V2 =
=
=
= 28.6
2
2
A2 π D2
s
 8 
π  ft 
 12 
Thus,
1
lb
1
slugs 
ft 
p2 = −γ z2 − ρV22 = −62.4 3 (6.0 ft) − 1.94 3   28.6 
2
2
s
ft
ft  
= −1168
lb
ft 2
= −8.11psi
2
Problem 3.80
Air is drawn into a small open-circuit wind tunnel as shown in the figure below. Atmospheric pressure is 98.7 kPa (abs) and the temperature is 27°C. If viscous effects are
negligible, determine the pressure at the stagnation point on the nose of the airplane. Also
determine the manometer reading, h, for the manometer attached to the static pressure tap
m
within the test section of the wind tunnel if the air velocity within the test section is 50 .
s
Test
section
Inlet nozzle
Diffuser
Q
Fan
h
Inlet
Water
Solution 3.80
Test
section
Inlet nozzle
Diffuser
Q
Fan
(1)
(2)
Inlet
h
(3)
Water
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
with p1 = 0, V1 = 0, z1 = z2 and V2 = 0
Thus, p2 = 0 (gage)
Also,
m
p3 V32
p1 V12
+
+ z1 =
+
+ z3 , with z1 = z3 and V3 = 50
s
γ 2g
γ 2g
Thus,
p3
γ
=−
or ρ =
1
p
V32
or p3 = − ρV32 where ρ =
RT
2
2g
98.7 ×103
N
m2

J 
 286.9 kg ⋅ K  (273 + 27) K


= 1.147
kg
m3
Hence,
2
1
kg  m 
N
p3 = 1.147 3  50  = 1434 2
2
m  s 
m
Since γ
2
γ
,,
N
p
m 2 = 0.146 m
p3 = γ H2O h or h = 3 =
γ 2O 9.80 ×103 N
m3
1434
Problem 3.81
Air flows through the device shown in the figure below. If the flowrate is large enough, the
pressure within the constriction will be low enough to draw the water up into the tube. Determine the flowrate, Q, and the pressure needed at section (1) to draw the water into section (2). Neglect compressibility and viscous effects.
Free jet
25 mm
Air
(2)
Q
(1)
0.3 m
50 mm
50 mm
Water
Solution 3.81
p2
γ
+
p V2
V22
+ z2 = 3 + 3 + z3 ,
γ 2g
2g
where z2 = z3 , A 2 V2 =A3V3
2
2
D 
 50 mm 
or V2 =  3  V3 = 
 V3 = 4V3
 25 mm 
 D2 
Thus, since p3 = 0
p2
γ
=
V32 V22
15V32
−
=−
2g 2g
2g
But
p2 = −γ w h
so that
p2
γ
γ
γ
=− w h=−
9.80 ×103
12
Thus,
−245 m = −
15V32
m

2  9.81 2 
s 

N
m3
N
m3 (0.3 m) = −245 m
or
V3 = 17.9
m
s
Thus,
Q = A3V3 =
π
m
m3

(0.050 m)2 17.9  = 0.0351
4
s 
s

Also,
p V2
p1 V12
A
+
+ z1 = 3 + 3 + z3 where V1 = 3 V3 = V3
γ 2g
γ 2g
A1
and z1 = z3
Thus,
(
)
1
p1 = p3 + γ ( z3 − z1 ) + ρ V32 −V12 or
2
p1 = 0
Problem 3.82
Water flows steadily from the large open tank shown in the figure below. If viscous effects
are negligible, determine (a) the flowrate, Q , and (b) the manometer reading, h.
4m
h
2m
Q
0.08 m
0.10 m
Mercury
Solution 3.82
(1)
4m
h
2m
Q
(2)
(3)
0.08 m
0.10 m
Mercury
(a) From the Bernoulli’s equation,
1
1
p1 + ρV12 + γ z1 = p2 + ρV12 + γ z2 , where p1 = p2 = 0, V1 = 0, z1=4 m, and z2 = 0
2
2
Thus,
1
2
1
2
γ z1 = ρV22 , or ρ gz1 = ρV22 so that V2 = 2 gz1
or
m
m

V2 = 2  9.81 2  (4 m) = 8.86
s
s 

Hence,
Q = A2V2 =
π
m
m3

(0.10 m)2  8.86  = 0.0696
4
s 
s

(b) From the Bernoulli’s equation,
1
1
p3 + ρV32 + γ z3 = p2 + ρV22 + γ z2 , where z2 = z3 and p2 = 0
2
2
so that
p3 =
(
1
ρ V22 −V32
2
)
2
2
D 
 0.1m 
A
m
m
Also, A2V2 = A3V3 so that V3 = 2 V2 =  2  V2 = 
 8.86 = 13.84
A3
s
s
 0.08 m 
 D3 
Hence,
2
2
1
kg  
m 
m 
N
p3 =  999 3   8.86  − 13.84   = −56500 2
2
s  
s  
m  
m
(1)
Also, from the manometer,

 0.08  
p3 = −γ Hg h + γ H 2O  2 m + 
m
 2  

N 
N 


= −  133 × 103 3  h +  9.80 × 103 3  (2.04 m)
m 
m 


= −133 ×103 h + 1.99 × 10 4
N
, where h
m
= −133 ×103 h + 1.99 × 10 4
N
m2
Thus, from Eqs. (1) and (2):
−5.65 × 10 4
N
m
or
h = 0.574 m
2
m2
(2)
Problem 3.83
Water from a faucet fills a 16-oz glass (volume = 28.9 in.3 ) in
. If the diameter of the jet
, what is the diameter of the jet when it strikes the water surface
leaving the faucet is
in the glass which is positioned
below the faucet?
Solution 3.83
p1 V12
p V2
+
+ z1 = 2 + 2 + z2
γ 2g
γ 2g
with p1 = p2 = 0, z1 = 14 in., z2 = 0
Q
(1)
Thus,
D1 = 0.60 in.

V2 
Q
υ
V2 = 2 g  z1 + 1  where V1 =
=

2 g 
A1 A1t

or
3
 1ft 
(28.9 in.3 ) 

12 in. 
ft

= 0.426
V1 =
2
s
π  0.60  2

 ft (20 s)
4  12 
Hence,
2

ft  

0.426


 
ft   14
ft
s 


ft +
V2 = 2  32.2 2 
= 8.67
ft  
s

s   12

2  32.2 2  


s 


But,
A1V 1= A 2V 2 so that D12V1 = D22V2
or
1
ft  2

 0.426 s 
 V1
D2 =   D1 = 
 (0.60 in.) = 0.132 in.
ft
 V2 
 8.76 
s 

1
2
(2)
Problem 3.84
The nozzle shown in the figure below has two water manometers to indicate the static presand
, respectively. Air flows
sures at sections 1 and 2. The diameters D1 and D2 are
through the nozzle, and the air and water temperatures are 60°F. Find the air volume flow
rate. Assume constant density and inviscid flow. Neglect elevation changes.
patm = 14.7 psia
D2
D1
h2 = 8.0 in.
h1 =
1
in.
32
Solution 3.84
Apply Bernoulli’s equation to a streamline connecting cross section 1 and cross section 2.
p1 V12
p V2
+
+ gz1 = 2 + 1 + gz2 .
ρ
2
ρ
2
Since z1 ≅ z2 and
V1 =
4Q
π D12
and
V2 =
4Q
π D22
to get
2
1  4Q   1
1  p1 − p2

  4 − 4  =
ρ
2  π   D2 D1 
Q=
or
π 2 ( p1 − p2 )
 1
1 
8ρ  4 − 4 
D

 2 D1 
Now
−
=ρ

=





=

lb
 62.4 lb 
patm − p2 = ρw gh2 = 
(8.0 in.) = 0.289 2 ,

3
in.
 1728 in. 
,
.
and
p1 − p2 = ( 0.00113 + 0.289 )
lb
2
in.
= 0.290
lb
in.2
Then
=





⋅



π 



−



⋅









or
Q = 4.09
ft3
s
DISCUSSION You might be tempted to write Bernoulli’s equation from the atmosphere
(V = 0) to cross section 1. This is not valid since the pressure at cross section 1 is greater
than the atmospheric pressure, which indicates that a fan was used to put energy into the
flow. Bernoulli’s equation does not apply to a flow stream with a fan or a turbine or other
device that changes its energy.
Problem 3.85
Air flows steadily through a converging–diverging rectangular channel of constant width as
shown in the figure below. The height of the channel at the exit and the exit velocity are H 0
and V0 respectively. The channel is to be shaped so that the distance, d, that water is drawn
up into tubes attached to static pressure taps along the channel wall is linear with distance
d

along the channel. That is d =  max  x , where L is the channel length and dmax is the max L 
imum water depth (at the minimum channel height: x = L). Determine the height, H ( x ), as
a function of x and the other important parameters.
x
x=L
x=0
L
V0
Q
Air
dmax
H(x)
H0
d
Water
Solution 3.85
1
1
p + zγ + ρV 2 = p0 + z0γ + ρV02 where ρ = air density
2
2
where
z = z0 , p0 = 0, p = −γ H 2O d = −γ H 2O
d max
x
L
Thus,
−γ H 2O
d max
1
1
x + ρV 2 = ρV02
L
2
2
But
AV = A 0V 0, or V =
A0
H
V0 = 0 V0
A
H
so that
2
−γ H2O
dmax
1 H
1

x + ρ  0 V0  = ρV02
2 H
2
L

or
H
=
H0
1
 2γ H2O d max  x
1+ 


ρV02

L
Typical shapes are shown below.
H/H0 vs x/L
1
0.8
2 γ H O dmax
2
4
0.6
ρ V 02
H/H0
2
1
0.4
0.2
0
1
0.8
0.4
0.6
x/L
0.2
0
Problem 3.86
Water flows from a large tank and through a pipe of variable area as shown in the figure
  x 
1 −  1 −  


, where A0 is the area at the bebelow. The area of the pipe is given by A = A0 
2




ginning ( x = 0) and end ( x = ) of the pipe. Plot graphs of the pressure within the pipe as a
function of distance along the pipe for water depths of h = 1, 4,10, 25 m.
h
ℓ
Free jet
x
x=0
Solution 3.86
(1)
h
ℓ
Free jet
A0
x
(2)
x=0
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 where p1 = p2 = 0, V1 = 0, z1 = h , and z2 = 0
γ 2g
γ 2g
Thus,
V2 = 2 gh Also, at any x location AV = A2V2 so that with A2 = A0
V=
A0
V2
A
and
2
1
1 A 
p1 V12
p V2
+
+ z1 = +
+ z, or p = γ h − ρV 2 = γ h − ρ  0  (2 gh )
2
2  A
γ 2g
γ 2g
Thus,
  A 2 
p = γ h 1 −  0   ,
  A  
or with γ = 9.80
kN
m3





 kN
1
p = 9.80h 1 −
, where h
2  2
x  x   m
 
 1 − 2 1 −   
 
 
(1)
m
x
x
Plot p = p   for 0 ≤ ≤ 1 with h = 1, 4,10, and 25 m
 
x
ℓ
0
0
0.2
0.4
0.6
0.8
h=1m
–20
h=4m
h = 10 m
p, kN –40
2
m
–60
h = 25 m
1.0
Problem 3.87
If viscous effects are neglected and the tank is large, determine the flowrate from the tank
shown in the figure below.
Oil,
SG = 0.81
2m
0.7 m
Water
50-mm
diameter
Solution 3.87
(0)
Oil,
SG = 0.81
h=2m
γ0
(1)
(2)
0.7 m
γ
Water
50-mm
diameter
p1 V12
p2 V22
+
+z =
+
+ z where p1 = p0 + γ 0 h = γ 0 h
γ 2g 1 γ 2g 2
z1 = 0.7 m, z2 = 0, and V1 = 0
Thus,
γ h

γ 0h
γ
V2
+ z1 = 2 or V2 = 2 g  0 + z1  where 0 = 0.81
γ
2g
γ
 γ

and
Q = A2V2 =
π
4
D22V2
Thus,
Q=
π
m
m3

(0.050 m)2 2  9.81  ( 0.81( 2 m ) + 0.7 m ) = 0.0132
4
s 
s

Problem 3.88
Water flows steadily downward in the pipe shown in the figure below with negligible losses.
Determine the flowrate.
Open
Oil SG = 0.7
1.5 m
1.2 m
2m
1m
Solution 3.88
From the Bernoulli’s equation,
(1)
p1
γ
+ z1 +
Open
V12 p2
V2
=
+ z2 + 2 where z1 − z2 = 2 m
2g γ
2g
γ
h1
and
AV
1 1 = A2V2 or
Oil SG = 0.7
π
4
2
(1.2 m) V1 =
π
4
(1)
2
(1m) V2
1.2 m
or
(2) V1 = 0.694V2
2 m = h2
1m
(2)
Also, from the manometers,
p1 = γ h1 and p2 = γ oil h1 + γ h2 , where γ oil = 0.7γ
Thus,
p2 − p1 = γ (0.7h1 + h2 ) − γ h1
or
1.5 m = h1
(3)
p2 − p1
γ
= h2 − 0.3h1 = 2 m − 0.3(1.5 m) = 1.55 m
Now, from Eq. (1),
z1 − z2 =
p2 − p1 V22 V12
+
−
γ
2g 2g
which, when combined with Eqs. (2) and (3), gives:
2 m = 1.55 m +
V22
m

2  9.81 2 
s 

(1 − (0.694)2 )
or
V2 = 4.13
m
s
Hence,
π
m
m3
Q = A2V2 = (1m)  4.13  = 3.24
4
s 
s

2
Problem 3.89
Water flows steadily from a large open tank and discharges into the atmosphere through a
3-in.-diameter pipe as shown in the figure below. Determine the diameter, d, in the narrowed
section of the pipe at A if the pressure gages at A and B indicate the same pressure.
B
9 ft
3-in. diameter
8 ft
A
16 ft
diameter = d
Solution 3.89
B
(3)
9 ft
3-in. diameter
(1)
8 ft
A
(4)
16 ft
diameter = d
(2)
1
1
p4 + ρV42 + γ z4 = p2 + ρV22 + γ z2 , where z2 = 0 and p2 = 0
2
2
Thus, since p3 = p4
1
1
p3 + ρV42 + γ z4 = ρV22
2
2
(1)
However,
1
1
p1 + ρV12 + γ z1 = p2 + ρV22 + γ z2 where p1 = p2 = V1 = z2 = 0
2
2
So that
1
ρV22 = γ z1 or
2
V2 = 2
1
2
 
γ
ft 
ft
z1 = 2 gz1 =  2  32.2 2  (16 ft)  = 32.1
ρ
s
s 
 

But
1
1
p3 + ρV32 + γ z3 = p2 + ρV22 + γ z2 where V2 = V3 since A2 = A3
2
2
Thus,
lb 
lb

p3 = −γ z3 = −(16 + 9) ft  62.4 3  = −1560 2
ft 
ft

(2)
From Eqs. (1) and (2):
2
1
slugs 
1
slugs 
ft 
−1560 2 + 1.94 3 V42 = 1.94 3   32.1 
2
2
s
ft
ft 
ft  
lb
or
V4 = 46.1
ft
s
Since A4V4 = A2V2 it follows that
π
4
d 2V4 =
π
4
D22V2
or
d = D2
ft
V2
s = 2.50 in.
= (3 in.)
ft
V4
46.1
s
32.1
Problem 3.90
Water flows from a large tank as shown in the figure below. Atmospheric pressure is
, and the vapor pressure is
. If viscous effects are neglected, at what
height, h, will cavitation begin? To avoid cavitation, should the value of D1 be increased or
decreased? To avoid cavitation, should the value of D2 be increased or decreased? Explain.
h
D3 = 4 in.
D1 = 1 in.
D2 = 2 in.
Solution 3.90
(0)
h
D3 = 4 in.
(2)
(1)
D1 = 1 in.
D2 = 2 in.
p0 V02
p1 V12
+
+z = +
+ z where p0 = 14.5 psia, p1 = 1.60 psia, z0 = h, z1=0, and V0 = 0
γ 2g 0 γ 2g 1
Thus,
h=
p1 − p0 V12
+
γ
2g
(1)
However,
2
 D2 
AV
 V2
1 1 = A2V2 or V1 = 
D
 1
where
p0 V02
p2 V22
+
+z =
+
+ z with p0 = p2 and z2 = 0
γ 2g 0 γ 2g 2
Thus,
V22
=h
2g
so that
4
 D2  2
4

 V2
2
D1 
 D2 
V1

=
=
 h
2g
2g
 D1 
(2)
Combine Eqs. (1) and (2) to obtain
h=
p1 − p0
γ
4
D 
+ 2  h
 D1 
or
h=
p0 − p1
 D  4 
γ  2  − 1
 D1 



lb 
in.2 
144


in.2 
ft 2 
= 1.98 ft
4


lb  2 in. 
62.4 3 
 − 1
ft  1in. 

(14.5 − 1.60)
=
(3)
From Eq. (3) it is seen that h increases in increasing D1 and decreasing D2. Thus, to avoid
cavitation (i.e., to have h small enough) D1 should be increased and D2 decreased.
Problem 3.91
gal
. If the drain is closed,
min
the water will eventually flow through the overflow drain holes rather than over the edge of
the sink. How many 0.4-in.-diameter drain holes are needed to ensure that the water does
not overflow the sink? Neglect viscous effects.
Water flows into the sink shown in the figure below at a rate of 2
1 in.
Q = 2 gal/min
0.4-in.-diameter
holes
Stopper
Solution 3.91
1 in.
Q = 2 gal/min (1)
(2)
0.4-in.-diameter
holes
Stopper
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = 0, V1 = 0, and z2 = 0,
γ 2g
γ 2g
Thus,
1
 
ft  1 + 0.2   2
ft
V2
ft   = 2.54
z1 = 2 or V2 = 2 gz1 = 2  32.2 
2g
s  12
s

 
p2 = 0
dj
dh
CC = 0.61
(a) Knife edge
CC = 1.0
(b) Well rounded
CC = A j /A h = (dj /dh)2
CC = 0.50
CC = 0.61
(c) Sharp edge
(d) Re-entrant
Also,
Q = nA2V2 = nCc
π
4
d22V2 ,
where n = number of holes required, d2 = 0.4 in.,
and C c = contraction coef. = 0.6l (see the figure above)
Thus, with
Q=2
3
gal  1min   231in.3  1ft3 
−3 ft
=
×
4.46
10
,





3
min  60 s   1gal 
s
1728
in.



ft3 
4  4.46 ×10 −3


s 
4Q

=
= 3.30
n=
2
π Cc d22V2
ft 
 0.4  2 
π (0.61) 
 ft  2.54 
s
 12 

Thus, 4 holes are needed.
Problem 3.92
What pressure, p1, is needed to produce a flowrate of 0.09
figure below?
p1
Air
Gasoline
2.0 ft
Salt water
SG = 1.1
3.6 ft
0.06-ft diameter
Solution 3.92
p1
Air
Gasoline
(2)
2.0 ft = h
Salt water
SG = 1.1
3.6 ft
0.06-ft diameter
(3)
p3 V32
V22
+
+z =
+
+ z where p2 = p1 + γ 0 h,
γ 2g 2 γ 2g 3
p2
z2 = 3.6 ft,
z3 = 0 and V2 = 0
Thus,
p1 + γ 0 h
γ
V32
+ z2 =
2g
where Q = AV
3 3=
π
4
D32V 3
p3 = 0
ft3
from the tank shown in the
s
or

ft3 
4  0.09


s 
4Q
ft

=
= 31.8
V3 =
2
2
s
π D3
π (0.06 ft)
Thus,
2


ft 
31.8



 V32

 
lb    
lb
s
p1 = γ 
− z2  − γ 0 h = 1.1 62.4 3  
− 3.6 ft  − 42.5 3 (2.0 ft)
 2g


ft    2  32.2 ft 
ft
 


 

2
s 
 

or
p1 = 746
lb
ft 2
= 5.18 psi
Problem 3.93
The vent on the tank shown in the figure below is closed and the tank pressurized to increase the flowrate. What pressure, p1, is needed to produce twice the flowrate of that when
the vent is open?
p1
Vent
4 ft
Air
10 ft
Water
Q
Solution 3.93
p1
Vent
4 ft
Air
(1)
10 ft
Water
Q
(2)
With the vent open:
p1 V12
p2 V22
+
+z =
+
+ z where p1 = p2 = 0, V1 = 0, z2 = 0
γ 2g 1 γ 2g 2
Thus,
z1 =
ft 
ft
V22

or V2 = 2 gz1 = 2  32.2 2  (10 ft) = 25.4
s
2g
s 

To have double the flowrate with the vent closed ( p1 ≠ 0) :
p1
γ
+ z1 =
V22
ft 
ft

where for this case V2 = 2  25.4  = 50.8
2g
s
s

Thus,
2
ft 

50.8 

p1
s
+ 10.4 ft= 
lb 
ft 


2  32.2 2 
 62.4 3 
ft 
s 


or
p1 = 1876
lb
ft 2
= 13.0 psi
Problem 3.94
A sump pump is submerged in 60°Fordinary water that vaporizes at a pressure of 0.256 psia.
The pump inlet has an inside diameter of 2.067 in. and is 15ft below the water surface. Find
the maximum possible flow rate before cavitation (vaporization) occurs.
Solution 3.94
Assume inviscid flow and constant water density and apply Bernoulli’s equation between
points 1 and 2.
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2 .
ρ
2
ρ
2
Now V1 ≈ 0 , p1 = p atm , p2 = 0.256 psia , and z1 − z2 = h
Then
V2 =
2 ( p1 − p2 )
ρ
+ 2 g ( z1 − z2 ) .
The numerical values give
2 (14.7 − 0.256 )
V2 =
lb 
in2 
144


in2 
ft 2 
slugs   lb ⋅ s2 

1.938



ft3   ft ⋅ slug 

ft 
ft

+ 2  32.2 2  (15 ft ) = 55.8
s
s 

The flow rate is
Q = A2V2 =
ft 3
Q = 1.30
s
π
4
D22V2 =
π
4


ft 
55.8 

2 
s
 144 in  
( 2.067 in.)2 
ft 2
Problem 3.95
Water is siphoned from a large tank and discharges into the atmosphere through a
2-in.-diameter tube as shown in the figure below. The end of the tube is 3 ft below the tank
bottom, and viscous effects are negligible. (a) Determine the volume flowrate from the
tank. (b) Determine the maximum height, H , over which the water can be siphoned without
cavitation occurring. Atmospheric pressure is 14.7 psia, and the water vapor pressure is
0.26 psia.
H
2-in. diameter
9 ft
3 ft
Solution 3.95
(3)
H
(1)
2-in. diameter
9 ft
3 ft
(2)
(a) From the Bernoulli’s equation,
p1 V12
p2 V22
+
+z =
+
+ z , where p1 = p2 = 0 and V1 = 0.
γ 2g 1 γ 2g 2
Thus,
z1 =
V22
+ z2
2g
or
ft 
ft

V2 = 2 g(z1 − z2 ) = (2)  32.2 2  (9ft + 3ft) = 27.8
s
s 

Hence,
π2
2
ft 
ft3
 
Q = A2V2 =  ft   27.8  = 0.607
4  12  
s
s
(b) From the Bernoulli’s equation,
p3 V32
p V2
+
+ z3 = 2 + 2 + z2 , where V2 = V3 since Q = A2V2 = A3V3 and A2 = A3
γ 2g
γ 2g
Thus, with z3 − z2 = H + 9ft + 3ft = H + 12 ft ,
p3 + γ ( z3 − z2 ) = p2
where p2 = 14.7 psia and p3 = 0.26 psia
Hence,
lb 
lb 144 in.2

+
=
−
62.4
H
12
ft
(14.7
0.26)
(
)


ft 3 
in.2 ft 2

or
H = 21.3 ft
Problem 3.96
Water flows in the system shown in the figure below. Assume frictionless flow. (a) Calculate
the rate Q at which water must be added at the inlet to maintain the 16-ft height.
(b) Calculate the height h in feet of water in the static-pressure tube.
1
16 ft
Q
2
A = 0.4 ft 2
h
A = 0.1 ft 2
3
4
Solution 3.96
(a) Assume constant fluid density and apply Bernoulli’s equation from the free surface 1 to
the jet discharge 4.
p1 V12
p V2
+
+ gz1 = 4 + 4 + gz4 .
ρ
2
ρ
2
Now p1 = p4 = p atm , V1 ≈ 0 , and z1 − z4 = 16 ft . Then
ft 
ft

V4 = 2 g ( z1 − z4 ) = 2  32.2 2  (16 ft ) = 32.1 .
s
s 

To maintain the constant water depth, Qin = Qout = Q = V4 A4 so
(
ft 

Q = V4 A4 =  32.1  0.1ft 2
s

)
ft3
Q = 3.21 .
s
(b) Assume constant water density and apply Bernoulli’s equation between the free water
surface 1 and points 3.
p V2
p 1 V12
+
+ gz1 = 3 + 3 + gz3 .
ρ
ρ
2
2
Now p1 = 0 ( psig ) , V1 ≈ 0 , and set z3 = 0 . This gives
g z1 =
p3 V32
+
.
ρ
2
From part (a), Q = 3.21
ft 3
s
so
ft 3
Q
s = 8.025 ft
V3 =
=
A3 0.4 ft 2
s
3.21
The pressure p3 is found from the monometer rule,
p3 = ρ gh
so Bernoulli's equation gives
2
g z1 = gh +
h = 15ft
V32
,
2
ft 

8.025 
2

V
s
h = z1 − 3 = 16 ft − 
ft 
2g

2  32.2 2 
s 

Problem 3.97
Water flows steadily from the pipe shown in the figure below with negligible viscous effects.
Determine the maximum flowrate if the water is not to flow from the open vertical tube
at A.
A
3 ft
Q
0.15-ft diameter
0.1-ft diameter
End of pipe
Solution 3.97
A
3 ft
Q
0.15 ft diameter
0.1 ft diameter
(1)
p1 V12
p V2
p 
0.15 
+
+ z1 = 2 + 2 + z2 , where z1 = z2 and 1 =  3 +
 ft,
γ 2g
γ 2g
2 
γ 
Thus,
V12 V22
3.075 ft +
=
2g 2g
But since ρ1 = ρ2 , then AV
1 1 = A2V2 or
(1)
π
4
π
( 0.15ft )2 V1 = ( 0.10 ft )2 V1
4
so that
V2 = 2.25V1
(2)
By combining Eqs.(1) and (2),
2
2 ft
V1 2
ft 2
s
s2
3.075 ft +
=
ft 
ft 


2  32.2 2  2  32.2 2 
s 
s 


(
2.25V12
)
or
V1 = 6.98
ft
s
Thus, Q = AV
1 1=
π
ft 
ft3

(0.15 ft)2  6.98  = 0.123
4
s
s

Problem 3.98
JP-4 fuel (SG = 0.77) flows through the Venturi meter shown in the figure below with a
ft
velocity of 15 in the 6-in. pipe. If viscous effects are negligible, determine the elevation, h,
s
of the fuel in the open tube connected to the throat of the Venturi meter.
6 ft
6 in.
JP-4 fuel
20°
8 in.
4 in.
h
6 in.
V = 15 ft/s
Solution 3.98
6 ft
6 in.
(2)
JP-4 fuel
8 in.
30°
4 in.
(1)
h
6 in.
V = 15 ft/s
8
ft
p1 V12
p2 V22
+
+ z1 =
+
+ z2 , where z1 = 0, z2 = ft, and V2 = 15
12
s
γ 2g
γ 2g
Also, AV
1 1 = A2V2
or
2
2
D 
 6 in.   ft 
A
ft
V1 = 2 V2 =  2  V2 = 
 15  = 33.75
A1
s
 4 in.   s 
 D1 
(1)
Thus, with
p2
γ
= 6 ft Eq. (1) becomes
2
2
ft 

 ft 
33.75 

 15 
p1 
8
s
s
+
= 6 ft + 
+ ft
ft 
ft  12
γ


2  32.2 2 
2  32.2 2 
s 
s 


or
p1
γ
= −7.53ft
But
p1
γ
= −h so that h = 7.53ft
Problem 3.99
Water, considered an inviscid, incompressible fluid, flows steadily as shown in the figure
below. Determine h.
Air
h
Q = 4 ft3/s
Water
0.5-ft diameter
1-ft diameter
3 ft
Solution 3.99
Air
h
ℓ
Q = 4 ft3/s
Water
(2)
0.5-ft diameter
1-ft diameter
3 ft
(1)
1
1
p1 + γ z1 + ρV12 = p2 + γ z2 + ρV22
2
2
where z1 = 0 , z2 = 3ft , V2 = 0 , and V1 =
4
ft3
s
Q
ft
=
= 5.09
A1 π 1ft 2
s
( )
4
Thus,
2
1
slugs 
ft 
lb
p1 + 1.94 3   5.09  = p2 + 62.4 3 (3ft)
2
s
ft  
ft
or
p1 − p2 = 162
lb
ft 2
(1)
But from the manometer,
p1 − γ ( + 3ft) + γ ( h + ) = p2
or
p1 − 62.4
lb
ft
3
(3 ft) + 62.4
lb
ft 3
h = p2
Hence,
p1 = p2 + 187 − 62.4h which when combined with Eq. (1) gives
p2 + 187 − 62.4h − p2 = 162
or
h = 0.400 ft
Problem 3.100
Determine the flowrate through the submerged orifice shown in the figure below if the contraction coefficient is Cc = 0.63 .
6 ft
4 ft
3-in.
diameter
2 ft
Solution 3.100
(1)
6 ft
(2)
(3)
3-in.
diameter
2 ft
4 ft
p1 V12
p2 V22
p
+
+ z1 =
+
+ z2 , where p1 = 0, V1 = 0, z1 = 4 ft, z2 = 0, and 2 = 2 ft
γ 2g
γ 2g
γ
Thus,
4 ft=2 ft +
V22
ft 

2  32.2 
s

or
V2 = 11.34
ft
s
so that
π3
2
ft 
ft3
 
Q = A2V2 = Cc A3V2 = (0.63)  ft  11.34  = 0.351
4  12  
s
s
Problem 3.101
The water clock (clepsydra) shown in the figure below is an ancient device for measuring
time by the falling water level in a large glass container. The water slowly drains out
of a container
through a small hole in the bottom. Determine the approximate shape
of circular cross section required for the water level to fall at a constant rate of 5 cm / hr if
in diameter. What size container ( h and R0 ) permits the clock to
the drain hole is
run for 24 hr without refilling?
R0
Water
level
R (z)
z
h
Solution 3.101
Assume constant fluid density and inviscid flow and apply Bernoulli’s equation between the
surface (0) and the exit hole (1) where z1 = 0.
V02
p1 V12
+
+ gz0 = +
.
ρ
2
ρ
2
p0
Since p0 = p1 = p atm and V0 ≅ 0 ,
V1 = 2 gz .
(1)
Applying the continuity equation. to the clepsydra gives
m out =
where
d ( mcv )
dt
or
ρV1A 1= − ρπ R2
dz
dt
dz
π D12
is negative. Substituting A1 =
and Eq. (1) gives
dt
4
π D12
4
2 gz = −π R 2
dz
.
dt
Solving for R gives
1
4



gz 
.
R = D1 
  dz 2 
8  
  dt  
Substituting the known numerical values,
1
2 4

m
s 

  9.81 2  z (m)  3600  
hr  

s 
R = 0.0015 m  
2


m



8  0.05 
hr 



1
R = 0.424 z 4 with R and z in meters.
The depth of the clepsydra to run 24 hours is
 cm 
h = 24 hr  5

 hr 
or
h = 120 cm = 1.20 m
Then
R0
1
= R ( z = h ) = 0.424(1.2) 4
R0 = 0.444 m .
Problem 3.102
A long water trough of triangular cross section is formed from two planks as is shown in
remains at the junction of the two planks. If the water
the figure below. A gap of
, how long a time does it take for the water depth to reduce to
depth initially was
90°
2 ft
0.1 in.
Solution 3.102
ℓ
(1)
h
(2)
w
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
(1)
where p1 = 0, p2 = 0, z1 = h, and z2 = 0
Also, V1A1 = V2 A2 or since
w it follows that V1
Thus, Eq. (1) gives
V2 = 2 gh so that
− A1
dh
= A2 2 gh
dt
with A1 = b = 2 bh and A2 = bw
where b is the tank length.
Thus,
−2bh
dh
= bw 2 gh
dt
V2 , where V1 = −
dh
dt
?
or
hdh = −w
g
dt which can be integrated to give
2
h f =1 1
h 2 dh = −w

hi =2
tf
g
dt
2 t
i =0
or
2
tf =
3w
3
3
 3 3 3
2 2
2
2
 hi − h f 2  =
 2 2 − 12  ft 2 = 36.5 s
ft 
g
 0.1 



 3
 ft 32.2 2 
s
 12 
Problem 3.103
Pop (with the same properties as water) flows from a 4-in. -diameter pop container that
contains three holes as shown in the figure below. The diameter of each fluid stream is
, and the distance between holes is
If viscous effects are negligible and quasi-
steady conditions are assumed, determine the time at which the pop stops draining from the
top hole. Assume the pop surface is 2 in. above the top hole when t = 0.
Surface at t = 0
2 in.
0.15 in.
2 in.
2 in.
4 in.
Solution 3.103
Surface at t = 0
Surface
at t > 0
(1)
0.15 in. h1 = h
2 in.
h3
h2
2 in.
2 in.
(2)
(3)
4 in.
pi Vi 2
VT2
+
+ zT = +
+z
γ
2g
γ 2g i
pT
where ( i = 1,2,3), pT = pi = 0 ,
VT2
Vi 2 , and zT − zi = hi .
Thus Vi = 2 ghi .
dh
,
dt
With VT =
dh
dt
Q = Q1 + Q2 + Q3 = −AT
=
where
π 4
and A1 = A2 = A3 =
=
π  0.15 
2
ft  = 1.227 ×10−4 ft 2

4  12 
2

AT =  ft  = 0.0873ft 2
4  12 
Thus,
dh
2 gA1  h1 + h2 + h3  = −AT
, where h1 = h, h2 = h + L, h3 = h + 2L and L = 2 in.
dt
Hence,
t
0

A1 
dh
−  2g
  dt = 
AT  0
h + h + L + h + 2L

L
where t is the time it takes for the free surface to reach the upper hole ( h = 0),
or
L
AT
t=
A1 2 g 0
=
dh
( h + h + L + h + 2L )
L
0.0873ft 2

1
 
ft   2 0
(1.227 ×10−4 ft 2 ) (2)  32.2 2  


dh
( h + h + L + h + 2L )
s 
Thus,
L
t = 88.7 
0
dh
( h + h + L + h + 2L )
where L =
2
ft=0.1667 ft
12
Note: With L in feet, this equation gives t in seconds.
The numerical valve of the integral is obtained by using the trapezoidal rule (or the 1/ 3
Simpons Rule) since the closed form analytical solution is not given in integral tables. The
EXCEL spread sheet used for the trapezoidal rule this is given below.
L
t = 88.7  f ( h ) dh where f ( h ) =
0
1
( h + h + L + h + 2L )
 1 20

s 
≈ 88.7   ( fi + fi +1 )( hi +1 − hi )  = (88.7
) 0.120 ft  =10.7 s
2
ft
 i =1

h, (in.)
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
h, (ft)
f(h), (1/ft)1/2
0.0000
1.015
0.0083
0.914
0.0167
0.870
0.0250
0.837
0.0333
0.810
0.0417
0.786
0.0500
0.764
0.0583
0.745
0.0667
0.728
0.0750
0.712
0.0833
0.697
0.0917
0.684
0.1000
0.671
0.1083
0.659
0.1167
0.647
0.1250
0.637
0.1333
0.627
0.1417
0.617
0.1500
0.608
0.1583
0.599
0.1667
0.591
Sum of column = integral =
Thus, t = 88.7*0.12011 = 10.7 s
(1/2)*(fi + fi+1)*
(hi+1 − hi), (ft1/2)
i
0.00804
0.00743
0.00711
0.00686
0.00665
0.00646
0.00629
0.00614
0.00600
0.00587
0.00575
0.00564
0.00554
0.00544
0.00535
0.00526
0.00518
0.00510
0.00503
0.00496
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
0.12011
Problem 3.104
A spherical tank of diameter D has a drain hole of diameter d at its bottom. A vent at the
top of the tank maintains atmospheric pressure at the liquid surface within the tank. The
flow is quasi-steady and inviscid and the tank is full of water initially. Determine the water
depth as a function of time, h = h(t ), and plot graphs of h(t ) for tank diameters of 1, 5,10,
and 20 ft if d = 1in.
Solution 3.104
(1)
r
D
2
h
(2)
d
dh
p1 V12
p2 V22
+
+ z1 =
+
+ z2 , where p1 = 0, p2 = 0, z1 = h, z2 = 0 and V1 = −
dt
γ 2g
γ 2g
V2 if r
Thus,
V2 = 2 gh which when combined with AV
1 1 = A2V2 gives
−A1
dh
dh π 2
= A2 2 gh or −π r 2
= d 2 gh
dt
dt 4
where R2 = r2 + ( h − R )
with R =
(1)
2
r
D
= radius of tank
2
Thus, r = R 2 − ( h − R )2 so that Eq. (1) becomes
d2
2  dh
2

− R − (h − R)
=

 dt
4
2 gh
or
1
 3
d 2 2g
 h 2 − 2Rh 2  dh =
dt


4


h–R
D= R
2
d
which can be integrated from the initial time and depth (t = 0, h = 2R ) to an arbitrary time
and depth (t, h ) as
1
h  3
d2
 h 2 − 2Rh 2 dh =



2R 
2g
4
t
 dt
0
or
5
5
3
3
d 2 2g
2 2
4 
 h − (2R ) 2  − R  h 2 − (2R ) 2  =
t
 3 

5
4




Use d =
(2)
1
ft and g = 32.2 ft / s 2 and plot h = h(t ) for values of R = 0.5, 2.5, 5 and 10 ft
12
Note: It is easier to solve Eq. (2) as t = t( h ) rather than h = h(t )
Note: The time taken to empty the tank, te , is obtained from Eq. (2) with h = 0 as
te =
5
64R 2
15d 2 g
Results of an EXCEL Program to calculate h(t) from Eq. (2):
D = 1 ft
t, s
h, ft
0.00
1.000
0.09
0.950
0.35
0.900
0.77
0.850
1.34
0.800
2.05
0.750
2.89
0.700
3.84
0.650
4.91
0.600
6.06
0.550
7.30
0.500
8.60
0.450
9.94
0.400
11.31
0.350
12.69
0.300
14.06
0.250
15.37
0.200
16.61
0.150
17.72
0.100
18.62
0.050
19.14
0.000
D = 5 ft
t, s
h, ft
0
5.000
5
4.750
19
4.500
43
4.250
75
4.000
114
3.750
161
3.500
215
3.250
274
3.000
339
2.750
408
2.500
481
2.250
556
2.000
632
1.750
710
1.500
786
1.250
859
1.000
929
0.750
990
0.500
1041
0.250
1070
0.000
D = 10 ft
t, s
h, ft
0
10.00
28
9.50
110
9.00
242
8.50
422
8.00
647
7.50
913
7.00
1216
6.50
1552
6.00
1917
5.50
2308
5.00
2718
4.50
3143
4.00
3577
3.50
4014
3.00
4445
2.50
4862
2.00
5253
1.50
5603
1.00
5889
0.50
6053
0.00
D = 20 ft
t, s
h, ft
0
20
158
19
620
18
1370
17
2390
16
3661
15
5163
14
6876
13
8778
12
10,846
11
13,055
10
15,376
9
17,782
8
20,237
7
22,706
6
25,144
5
27,502
4
29,714
3
31,695
2
33,311
1
34,239
0
Below are the graphs of above results.
Water Depth vs Time
D = 5 ft
1.0
5
0.9
5
0.8
4
0.7
4
0.6
3
h, ft
h, ft
Water Depth vs Time
D = 1 ft
0.5
0.4
2
0.3
2
0.2
1
0.1
1
0
0.0
0
5
10
15
20
0
t, s
Water Depth vs Time
D = 10 ft
Water Depth vs Time
D = 20 ft
10
20
9
18
8
16
7
14
6
12
5
10
4
8
3
6
2
4
1
2
0
0
200 400 600 800 1000 1200
t, s
h, ft
h, ft
3
2000
4000
t, s
6000
0
0
10000 20000 30000 40000
t, s
Problem 3.106
When the drain plug is pulled, water flows from a hole in the bottom of a large, open cylindrical tank. Show that if viscous effects are negligible and if the flow is assumed to be quasisteady, then it takes 3.41 times longer to empty the entire tank than it does to empty the
first half of the tank. Explain why this is so.
Solution 3.106
Q = AV =
π
 dh 
d 2V = Atank  − 
4
 dt 
at t = 0
where
at t
V = 2 gh and Atank =
π
4
D2
H
Thus,
d
2
h
2
dh 
2 gh = D  − 
 dt 
D
d
or
V
2
dh
d
= −2 g   dt
h
D
Integrate from h = H at t = 0 to h at t:
h
2 t
dh
d 
 h = − 2 g  D   dt
0
H
or
2
h
d 
2 h  = − 2g   t
D
H
or
2
t=
H/2
2 D 
H − h 
 
2g  d  
Thus, to empty the tank,
2 D
t h =0 =
 
2g  d 
2
H
and to half empty the tank,
t
h=
H=
2
2
2
2 D 
H
2 D
1 

H
−
=

  
  H 1 −
2 
2g  d  
2g  d 
2 

Thus,
t0
t H
2
=
2 D
 
2g  d 
2 D
 
2g  d 
2
2
H
1 

H 1 −
2 

=
1
1 

1 − 2 


= 3.41
Problem 3.107
Someone siphoned 15 gal of gasoline from a gas tank in the middle of the night. The gas
tank measures 12in. wide × 24 in. long × 18 in. high and was full when the thief started. If
1
in. , find the minimum amount of time need2
ed to siphon the 15 gal from the tank. Assume that at any instant of time the steady-state
equations are adequate to predict the gasoline velocity in the siphon tube. Also assume that
the end of the siphon tube outside the gas tank is at the same level as the bottom of the
tank. You may consider the gasoline to be inviscid.
the siphoning tube has an inside diameter of
Solution 3.107
The minimum time is for inviscid flow. Apply Bernoulli’s equation from gasoline free surface (1) to siphon tube outlet (2). Note p1 = p2 = p atm and V0 ≈ 0 .
This gives
gz1 =
V22
+ gz2 .
2
or
V2 = 2 g ( z1 − z2 ) = 2 gh .
Applying the continuity equation to the gasoline in the tank gives
d ( mcv )
= min − mout = 0 − ρG A2V2 = ρ A 2 2 gh
dt
Now
d ( mcv ) d
dh
= ( ρG Ah ) = ρG A .
dt
dt
dt
The continuity equation gives
ρG A
dh
= − ρG A 2 2 gh
dt
or
h
t
A
dh
 h = −  A2 2 gdt
h
t =0
0
where h0 = 18 in. Integrating gives
A2
A
h − h0 = −
g
t
2
or
t=
Since A2 =
t=
2 A 


g  A2 
πd2
4
( h0 − h ) .
1
with d = in. ,
2
2  4A 
g  π d 2 
( h0 − h ) .
Now,

in.3 
15
gal
1728
(
) 

ft 3 
15 gal

h0 − h =
=
= 12.0 in.
gal 
A

(12 in. × 24 in.)  7.48 3 
ft 

or
h = h0 − 12.0 in. = 18.0 in. − 12.0 in. = 6.0 in.
The numerical values give




4 (12 in. × 24 in.) 
2  4A 
2
s

=
= 366


2
2

ft  
g πd 

ft
1 

 32.2 2   π  in. 
s  


2 
so

=

t = 190s.



(
−

) 




Problem 3.108
The surface area, A, of the pond shown in the figure below varies with the water depth, h, as
shown in the table. At time t = 0 a valve is opened and the pond is allowed to drain through
a pipe of diameter D. If viscous effects are negligible and quasisteady conditions are
assumed, plot the water depth as a function of time from when the valve is opened (t = 0)
until the pond is drained for pipe diameters of D = 0.5,1.0,1.5, 2.0, 2.5, and 3.0 ft . Assume
=
at t = 0.
Area A
h
3 ft
D
h (ft)
A [acres (1 acre = 43,560 ft2)]
0
2
4
6
8
10
12
14
16
18
0
0.3
0.5
0.8
0.9
1.1
1.5
1.8
2.4
2.8
Solution 3.108
dh
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 where p1 = 0, p2 = 0, z1 = h, z2 = −3ft and V1 = −
dt
γ 2g
γ 2g
Thus, V2 = 2 g ( h + 3) which when combined with AV
1 1 = A2V2 gives
−A1
dh π 2
= D 2 g ( h + 3) where A1 = A1 ( h ) as given.
dt 4
This can be rearranged and integrated to give
h
 A1
18ft
dh
π
=−
4
h+3
t
2 gD 2  dt = −
0
π
4
D 2 2 gt = −
π
4
D 2 2 × 32.2t
V2
or
t=
0.15q
D
2
18
dh
 A1 h + 3 , where t
s, A1
ft 2 , and h
ft
(1)
h
Note: It is easier to determine t as a function of h rather than h as a function of t
Note: t
D −2
An EXCEL Program using a trapezoidal integration approximation was used to calculate the results
shown below.
h, ft
18
16
14
12
10
8
6
4
2
0
A, acres
2.8
2.4
1.8
1.5
1.1
0.9
0.8
0.5
0.3
0
A, ft2
121,968
104,544
78,408
65,340
47,916
39,204
34,848
21,780
13,068
0
D = 0.5
ft
t, s
0
32,181
59,530
82,354
101,536
117,506
132,412
145,035
153,988
157,704
D = 1.0
ft
t, s
0
8045
14,882
20,589
25,384
29,377
33,103
36,259
38,497
39,426
D = 1.5
ft
t, s
0
3576
6614
9150
11,282
13,056
14,712
16,115
17,110
17,523
D = 2.0
ft
t, s
0
2011
3721
5147
6346
7344
8276
9065
9624
9857
D = 2.5
ft
t, s
0
1287
2381
3294
4061
4700
5296
5801
6160
6308
The graph for D = 1 ft is shown below. The shape of the curve is the same for any D
Water Depth vs Time
for D = 1 ft
20
18
16
14
h, ft
12
10
8
6
4
2
0
0
10,000
20,000
t, s
30,000
40,000
D = 3.0
ft
t, s
0
894
1654
2288
2820
3264
3678
4029
4277
4381
Problem 3.109
Water flows through a horizontal branching pipe as shown in the figure below. Determine
the pressure at section (3).
V3
A3 = 0.07 m2
(3)
V2
p2 = 350 kPa
A2 = 0.02 m2
(2)
(1)
V1 = 4 m/s
p1 = 400 kPa
A1 = 0.1 m2
Solution 3.109
Q1 = Q2 + Q3 or V3 =
m3
Q1 − Q2
2 m
=
0.1m
4
=
0.4
where Q1 = AV
1 1


s
A3
 s 
Also, Q2 = A2V2 where
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 with z1 = z2
γ 2g
γ 2g
Thus,
2
 m
4 
V22
400 kPa
350 kPa
s 
+ 
=
+
kN
kN
m
m
9.80 3 2  9.81  9.80 3 2  9.81 
2
m
m
s 
s2 


or
V2 = 10.78
m
s
Thus,
m3
m

− 0.2 m 2 10.78 
0.4
m
s
s 

= 2.63
V3 =
2
s
0.07 m
Then from
p V2
p1 V12
+
+ z1 = 3 + 3 + z3 with z1 = z3
γ 2g
γ 2g
we obtain
p3 = p1 +
γ
(V −V )
2g
2
1
2
3
kN
2
3
2
2 m
m
(4 − 2.63 ) 2
= 400 kPa+
m

s
2  9.81 2 
s 

or
p3 = (400 + 4.54)
kN
m2
= 404.5 kPa
9.80
Problem 3.110
The “wye” fitting shown in the figure below lies in a horizontal plane. The fitting splits the
inlet flow into two equal parts. At section 1, the water velocity is 12 ft / sec and the pressure
is 20 psig . Calculate the water pressure at sections 2 and 3. Assume inviscid flow and a
water temperature of 60°F.
½Q1
p2
D1
D2
D 2 = 3 in.
D3
D 3 = 4 in.
Q1
1
D 1 = 6 in.
p3
½Q1
Solution 3.110
Apply Bernoulli’s equation to a streamline between points 1 and 2. Assume no elevation
changes and constant water density.
p1 V12 p2 V22
+
=
+
.
ρ
2
ρ
2
Since Q =
π D2
4
V and Q2 = Q3 =
p2 = p1 −
ρ
2
(
V22 −V12
)
Q
,
2
8ρQ 2  1
1 
= p1 − 2 
−
.
π  4D24 D14 
In a similar manner, apply Bernoulli’s equation between points 1 and 3 to get
p3 = p1 −
8ρQ 2  1
1 
− 4 .
2 
4
π  4D3 D1 
The numerical values give
Q1 =
π D12
π6
2
ft3
  ft 
V1 =  ft  12  = 2.36
,
4
4  12   s 
s
2

lbm  
ft3  

8  62.4 3   2.36
 



s 
ft  

1
1


−
p2 = 20 psig −
4
4
2

ft ⋅ lbm  
in.   3   6  

π 2  32.2
144

  4  ft 

 ft  
lb ⋅ s2  
ft 2    12   12  

p2 = 17.1 psig ,
2

lbm  
ft3  

8  62.4 3   2.36
 



s 
ft  

1
1


−
p3 = 20 psig −
4
4
2

ft ⋅ lbm 
in.

 4 
6 
π 2  32.2
144 2   4  ft   ft  
2 


lb ⋅ s  
ft    12   12  

and
p3 = 19.7 psig.
Problem 3.111
Water flows through the branching pipe shown in the figure below. If viscous effects are
negligible, determine the pressure at section (2) and the pressure at section(3).
A3 = 0.035 m2
z3 = 10 m
Q1 = 1 m3/s
A1 = 0.1 m2
p1 = 300 kPa
z1 = 0
(3)
V2 = 14 m/s
A2 = 0.03 m2
z2 = 0
(2)
(1)
Solution 3.111
Along the streamline from (1) to (2):
p1 V12
p2 V22
+
+z =
+
+ z where z1 = z2 = 0 and
γ 2g 1 γ 2g 2
m3
Q
s = 10 m
V1 = 1 =
A1 0.1m 2
s
1
Thus,
2
2
N
 m
 m
300 × 10
 10 
 14 
2
p2
s 
m +  s  =
+ 
N
N
m
m
9.80 × 103 3 2  9.81  9.80 × 103 3 2  9.81 
2
m
m
s 
s2 


3
or
p2 = 2.52 × 105
N
m2
= 252 kPa
Along the streamline from (1) to (3):
p V2
p1 V12
+
+ z1 = 3 + 3 + z3
γ 2g
γ 2g
where since Q1 = Q2 + Q3 then
Q3 = A3V3 = Q1 − Q2 = Q1 − A2V2 so that
m3
 m
− 0.03 m 2 14 
Q − A2V2
s
 s  = 16.6 m
=
V3 = 1
2
s
A3
0.035 m
1
(1)
Thus, Eq. (1) becomes (with z1 = 0, z3 = 10 m )
2
2
m
N
 m

16.6 
300 × 10
 10 

2
p3
s 
m +  s  =
+
+ 10 m
m
m


3 N
3 N
9.80 × 10
2  9.81 2  9.80 × 10
2  9.81 2 
m3
m3
s 
s 


3
or
p3 = 1.14 × 105
N
m2
= 114 kPa
Problem 3.112
Water flows through the horizontal branching pipe shown in the figure below at a rate of
. If viscous effects are negligible, determine the water speed at section (2), the pressure
at section (3), and the flowrate at section (4).
A2 = 0.07 ft2
p2 = 5.0 psi
(2)
A3 = 0.2 ft2
V3 = 20 ft /s
(1)
(3)
A1 = 1 ft2
Q1 = 10 ft3/s
p1 = 10 psi
(4)
Solution 3.112
p1 V12
p2 V22
From (1) to (2):
+
+z =
+
+ z where z1 = z2 , p1 = 10 psi, p2 = 5psi ,
γ 2g 1 γ 2g 2
 ft3 
10

s 
ft
Q1

= 10
or V1 =
and V1 =
2
s
A1
(1ft )
Thus, with γ = ρ g
lb  
in.2   ft 2  lb  
in.2 

10
144
5
144




10




ft 2   s 
ft 2  V2 2
 in.2  
 in.2  
or
+
=
+
slugs 
slugs 
2
2


 1.94 3 
 1.94 3 
ft 
ft 


From (1) to (3):
V1 = 10
p V2
p1 V12
+
+ z1 = 3 + 3 + z3 where z1 = z3 ,
γ 2g
γ 2g
ft
ft
and V3 = 20
s
s
Thus,
2
2
lb  
in.2 

ft
ft




10
144



10 
 20 
ft 2 
 in.2  
p3
s
s

+
=
+ 
lb
lb
ft 
ft 


62.4 3
2  32.2 2  62.4 3 2  32.2 2 
ft
ft
s 
s 


p1 = 10 psi,
or
p3 = 1150
lb
ft 2
= 7.98 psi
Also,
Q4 = Q1 − Q2 − Q3 = Q1 − A2V2 − A3V3
or
Q4 = 10
ft3
ft 
ft3

 ft 
− 0.07 ft 2  29.0  − 0.2 ft 2  20  = 3.97
s
s
s

 s
Problem 3.113
Water flows from a large tank through a large pipe that splits into two smaller pipes as
shown in the figure below. If viscous effects are negligible, determine the flowrate from the
tank and the pressure at point (1).
3m
7m
0.03-m diameter
0.05-m diameter
(1)
0.02-m diameter
Solution 3.113
(0)
3m
7m
0.03 m
(2)
0.05 m
(3)
(1)
0.02 m
p0 V02
p2 V22
+
+z =
+
+ z where p0 = 0,
γ 2g 0 γ 2g 2
p2 = 0, V0 = 0, z0 = 7 m and z2 = 4 m
Thus,
m
m

V2 = 2 g − ( z0 − z2 ) = 2  9.81 2  (7 − 4) m = 7.67
s
s 

Similarly,
m
m

V3 = 2 g − ( z0 − z3 ) = 2  9.81 2  (7 m) = 11.7
s
s 

Thus, Q = Q2 + Q3 =
π
4
D22V2 +
π
4
D32V3
or
Q=
π
2
( 0.03 m )  7.67
4

3
m
m 
2
−3 m
+
0.02
m
11.7
=
9.10
×
10
(
)



s 
s  
s

Also,
−3 m
3
9.10 ×10
p0 V02
p V2
Q
s = 4.63 m
+
+ z0 = 1 + 1 + z1 where z1 = 0 and V1 =
=
π
γ 2g
γ 2g
A1
s
( 0.05 m )2
4
2

m 


 4.63  

V12 
N
s  
3 N 
p1 = γ  z0 −
7m − 
= 5.79 × 10 4 2
 = 9.80 × 10
3
m 
2 g 

m 
m

2  9.81 2  

s  


or
p1 = 57.9 kPa
Problem 3.114
A gutter running along the side of a house is 6 in. wide and 40ft long. During a hard
downpour, water is 1in. deep in the gutter. The gutter has only one down spout, and it is
3 in. in diameter. What is the velocity of the water entering the downspout? The pressure at
the downspout entrance is atmospheric pressure.
Solution 3.114
Use the following sketch.
Vg
d
g
h = 1.0″
Control
Volume
Vd
d
Apply the continuity equation to the above control volume. Assume constant water density
and uniform velocity across area Ag .
Vd
πd2
4
= AgVg
or
Vg =
π d 2Vd
4 Ag
(1)
 6 
where Ag = (40 ft)  ft  = 20 ft 2 .
 12 
Now apply Bernoulli’s equation from point ( g ) to point ( d ) .
pg Vg2
Vd2
+
+ gzd =
+
+ gzg
ρ
2
ρ
2
pd
Now set pg = 0 psig and pd = 0 psig .
Using Eqs. (1) and (2) gives
2
Vd2
1  π d 2Vd 
+ gzd = 
 + gz g
2
2  4 Ag 
or
Vd =
2 g ( z g − zd )
πd2 
1− 
 4 Ag 


2
(2)
The numerical values give
ft   1.0 

2  32.2 2  
ft 
12 

s


Vd =
2
  3 2 
 π  ft  
12  
1−  
 4(20 ft 2 ) 




or
Vd = 2.32
ft
s
Problem 3.115
Air, assumed incompressible and inviscid, flows into the outdoor cooking grill through nine
in.3
holes of 0.40-in. diameter as shown in the figure below. If a flowrate of 40
into the grill
s
is required to maintain the correct cooking conditions, determine the pressure within the
grill near the holes.
9 holes, each
0.40-in. diameter
Solution 3.115
in.3
3
s = 0.0231 ft and A = π D2
Q = 9A2V2 where Q =
2
2
s
4
in.3
1728 3
ft
40
Thus,

ft3 
4  0.0231 

s 
4Q
ft
Q

=
=
=
2.94
V2 =
2
9A2 9π D2 2
s
 0.4 
9π 
ft 
 12 
Also,
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 where p1 = 0, z1 = z2 , and V1 = 0
γ 2g
γ 2g
Thus,
2
1
1
slugs 
ft 
lb
p2 = − ρV22 = −  2.38 ×10−3 3   2.94  = −1.03 ×10−2 2
2
2
s
ft  
ft
or
p2 = −7.14 ×10−5 psi
Problem 3.116
An air cushion vehicle is supported by forcing air into the chamber created by a skirt
around the periphery of the vehicle as shown in the figure below. The air escapes through
the 3-in. clearance between the lower end of the skirt and the ground (or water). Assume
the vehicle weighs 10,000 lb and is essentially rectangular in shape, 30 by 65ft . The volume
of the chamber is large enough so that the kinetic energy of the air within the chamber is
negligible. Determine the flowrate, Q , needed to support the vehicle. If the ground clearance were reduced to 2 in. , what flowrate would be needed? If the vehicle weight were reduced to 5000 lb and the ground clearance maintained at 3 in. , what flowrate would be
needed?
Q
Fan
Vehicle
Skirt
3 in.
Solution 3.116
Q
Fan
Vehicle
Skirt
γ
(2)
(0)
3 in.
To support the load p0 =
W
where W = vehicle weight and A0 = (30 ft)(65 ft) = 1950 ft 2
A0
Also,
p0 V02
p V2
+
+ z0 = 2 + 2 + z2
γ 2g
γ 2g
where p2 = 0, V0 = 0, and z0 ≈ z2
so that
V2 =
2 p0
ρ
or V2 =
2W
Ao ρ
With h = ground clearance , it follows that
Q = A2V2 = 2h( L + b )V2
where L = 65 ft and b = 30 ft
Thus,
Q = 2h(65 ft + 30 ft)
2W
slug 

(1950 ft 2 )  2.38 ×10−3 3 
ft 

or
Q = 124.7 h W
Thus, if h =
if h =
ft3
where h
s
ft and W
lb
3
ft3
ft and W = 10,000 lb then Q = 3120
12
s
2
ft3
ft and W = 10,000 lb then Q = 2080
12
s
and if h =
3
ft3
ft and W = 5000 lb then Q = 2200
12
s
Problem 3.117
Water flows from the pipe shown in the figure below as a free jet and strikes a circular flat
plate. The flow geometry shown is axisymmetrical. Determine the flowrate and the manometer reading, H .
H
0.1-m
diameter
V
0.4 mm
0.2 m
0.01-m
diameter
Pipe
Q
Solution 3.117
H
0.1 m
diameter
(2)
(0)
V
0.4 mm
0.2 m
(1)
0.01 m
diameter
Pipe
Q
p1 V12
p2 V22
+
+z =
+
+ z where p1 = 0, p2 = 0, z1 = 0, and z2 = 0.2 m
γ 2g 1 γ 2g 2
Thus,
V12 V22
=
+ z2 where A1V 1= A 2V 2 = Q
2g 2g
or
(1)
V1 =
A2
π D2 h
4D h
4(0.1m)(4 ×10−4 m)
V2 =
V2 = 22 V =
V2 = 1.6V2
2
π 2
A1
D
(0.01m)
1
D1
4
Hence, Eq. (1) gives
m
m

(1.60V2 )2 = V2 2 + 2  9.81 2  (0.2 m) or V2 = 1.59
s
s 

so that
m
m

Q = A2V2 = π (0.1m)(4 × 10 −4 m) 1.59  = 2.00 × 10 −4
s 
s

Also,
p V2
p1 V12
+
+ z1 = 0 + 0 + z0 ,
γ 2g
γ 2g
where V0 = 0, z0 = 0.2 m, V1 = 1.60V2
m
m

or V1 = 1.60  1.59  = 2.54 , and p1 = 0
s 
s

Thus,
2
m

2.54 
2

p V
s 
− 0.2 m = 0.129 m
H = 0 = 1 − z0 = 
m
γ
2g

2  9.81 2 
s 

3
Problem 3.118
A conical plug is used to regulate the airflow from the pipe shown in the figure below. The
air leaves the edge of the cone with a uniform thickness of 0.02 m. If viscous effects are negm3
, determine the pressure within the pipe.
ligible and the flowrate is 0.50
s
V
Free jet
Pipe
0.20 m
3
Q = 0.50 m /s
0.23 m
Cone
0.02 m
V
Solution 3.118
V
Free jet
Pipe
0.20 m
Q = 0.50 m3/s
0.23 m
(1)
Cone
(2)
0.02 m
V
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where z1 = z2 and p2 = 0
γ 2g
γ 2g
Also,
V1 =
0.5
m3
s
Q
m
=
= 12.0
A1 π (0.23 m)2
s
4
and
m3
Q
Q
m
s
V2 =
=
=
= 19.9
A2 2π Rh 2π (0.2 m)(0.02 m)
s
0.5
Thus,
p1 =
1
1
kg 
m2
N
ρ (V22 −V12 ) = 1.23 3  (19.92 − 12.02 ) 2 = 155 2
2
2
m 
s
m
Problem 3.119
The figure below shows two tall towers. Air at 10 °C is blowing toward the two towers at
V0 = 30 km / hr . If the two towers are 10 m apart and half the air flow approaching the two
towers passes between them, find the minimum air pressure between the two towers. Assume constant air density, inviscid flow, and steady-state conditions. The atmospheric pressure is 101kPa .
d = 10 m
D = 60 m
D = 60 m
Top view
Elevation view
Solution 3.119
Apply Bernoulli’s equation to a streamline connecting the approach velocity V0 conditions
and a point between the cylinders (2, V2, p2). Assuming z0 = z2 ,
V02 p2 V22
+
=
+
ρ
2
ρ
2
p0
or
p2 = p0 +
ρ
2
(V −V )
2
0
2
2
Now the minimum pressure p0 occurs when V2 is a maximum value or when the distance
between the two towers is d. Then the volume flow rate is
Q = ( D + d )hV0 = dhV2
or
D+d 
V2 = 
V0
 d 
so Bernoulli’s equation is
p2 = p0 +
ρV02 
2
D+d  
1
−
 
 .
2   d  
Using Table B.4, the numerical values give
kg  
m

 1.25 3   30,000 
hr 
m 
p2 = 101kPa + 
2
s 

2  3600 
hr 

p2 = 98.9kPa
2


  70 m 2   N ⋅ s 2   kPa 
1 − 


  ∗
  10 m    kg ⋅ m   1000 N 



m2 
Problem 3.120
Water flows steadily from a nozzle into a large tank as shown in the figure below. The
water then flows from the tank as a jet of diameter d. Determine the value of d if the water
level in the tank remains constant. Viscous effects are negligible.
4 ft
0.1-ft diameter
0.15-ft diameter
3 ft
d
1 ft
Solution 3.120
4 ft
0.1-ft diameter
(2)
(1)
0.15-ft diameter
(4)
3 ft
d
1 ft
(3)
From the Bernoulli’s equation,
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where z1 = z2 and p2 = 0
γ 2g
γ 2g
Also,
p1
γ
= 4 ft and AV
1 1 = A2V2
or
π
4
D12V1 =
π
4
D22V2
(1)
so that
2
2
D 
 0.10 ft 
V1 =  2  V2 = 
 V2 = 0.444V2
 0.15ft 
 D1 
Thus, from Eqs. (1) and (2),
p1 V12 V22
+
=
γ 2g 2g
or
V22 −V12 ) (1 − (0.444)2 )V 2
(
2
=
4 ft =
ft 

2  32.2 2 
s 

2g
Hence,
V2 = 17.9
ft
s
so that
Q2 = A2V2 =
π
ft 
ft3

(0.10 ft)2  17.9  = 0.1407
4
s
s

Also,
ft 
ft

Q3 = Q2 where Q3 = A3V3 and V3 = 2 g (z4 − z3 ) = 2  32.2 2  (3ft − 1ft) = 11.35
s
s 

Hence,
π
ft 
ft3
d 11.35  = 0.1407
4 
s
s
2
or
d = 0.126 ft
(2)
Problem 3.121
A small card is placed on top of a spool as shown in the figure below. It is not possible to
blow the card off the spool by blowing air through the hole in the center of the spool. The
harder one blows, the harder the card “sticks” to the spool. In fact by blowing hard
enough, it is possible to keep the card against the spool with the spool turned upside down.
Give this experiment a try. (Note: It may be necessary to use a thumb tack to prevent the
card from sliding from the spool.) Explain this phenomenon.
Card
Spool
Q
Solution 3.121
As the air flows radially outward in the gap between the card and the spool, it slows down
since the flow area increases with r, the radial distance from the center.
That is,
rexit
Q = 2π rhV , or V =
Q
(see the figure).
2π hr
exit
h
r
V
(1)
If viscous effects are not important, then
p
V2
p V2
+
= constant = exit + exit
γ 2g
γ
2g
Q
or since pexit = 0 (a free jet) it follows that
2
1
1
 Q   1
2
2
−V 2 = 
p = ρ Vexit
−V 2 , where from Eq. (1) Vexit
  2 − 2
2
 2π h   rexit r 
(
)
But rexit > r so that p < 0 There is a vacuum within the gap.
The card is sucked against the spool. The harder one blows through the spool (larger Q ),
the larger the vacuum, and the harder the card is held against the spool.
Problem 3.122
Observations show that it is not possible to blow the table tennis ball from the funnel
shown in figure (a) below. In fact, the ball can be kept in an inverted funnel, figure (b) below, by blowing through it. The harder one blows through the funnel, the harder the ball is
held within the funnel. Try this experiment on your own. Explain this phenomenon.
Q
Q
(a)
(b)
Solution 3.122
From Bernoulli’s equation with negligible gravity,
1
p + ρV 2 = constant
2
Thus, large V means small p . In general, the flow field around the ball is as indicated.
V2 > V1
fairly large V
p < patm
(2)
free jet
p = patm
(1)
V1
(3)
Very low V
p ≈ patm
The net effect is a pressure force upward that can balance the weight of the ball.
Note: Can apply Bernoulli’s equation along streamline between (1) and (2), but not across
the streamlines from (1) to (3).
Problem 3.123
Water flows down the sloping ramp shown in the figure below with negligible viscous effects. The flow is uniform at sections (1) and (2). For the conditions given, show that three
solutions for the downstream depth, h2 , are obtained by use of the Bernoulli and continuity
equations. However, show that only two of these solutions are realistic. Determine these
values.
V1 = 10 ft/s
h2
h1 = 1 ft
H = 2 ft
V2
Solution 3.123
(1)
V1 = 10 ft/s
h2
h1 = 1 ft
(2)
H = 2 ft
V2
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = 0,
γ 2g
γ 2g
p2 = 0, z1 = 3 ft, and z2 = h2
(1)
Also, A1V 1= A 2V 2
or
 ft 
(1ft) 10 
h
 s  = 10
V2 = 1 V1 =
h2
h2
h2
Thus, Eq. (1) becomes
2
2
 10 
 ft 
10
 


 s  + 3ft =  h2 
+h
ft 
ft  2


2  32.2 2 
2  32.2 2 
s 
s 


or
64.4 h23 − 293h22 + 100 = 0
By using a root finding program, the three roots to this cubic equation are found to be:
h2 = 0.630 ft
h2 = 4.48ft
or
h2 = a negative root . Clearly, it is not possible (physically) to have h2 < 0
Thus, h2 = 0.630 ft or h2 = 4.48 ft
Problem 3.124
Water flows in a rectangular channel that is 2.0 m wide as shown in the figure below. The
upstream depth is 70 mm . The water surface rises 40 mm as it passes over a portion where
the channel bottom rises 10 mm . If viscous effects are negligible, what is the flowrate?
100 mm
70 mm
Q
10 mm
Solution 3.124
(2)
(1)
70 mm
Q
100 mm
10 mm
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1 = 0,
γ 2g
γ 2g
p2 = 0, z1 = 0.07 m,
and z2 = (0.01 + 0.10) m = 0.11m
Also,
A1V 1= A 2V 2
or
V2 =
h1
0.07 m
V1 =
V1 = 0.7V1
0.10 m
h2
Thus, Eq. (1) becomes
m
m

[1 − 0.7 2 ]V12 = 2  9.81 2  (0.11 − 0.7) m or V1 = 1.24
s
s 

Hence,
m
m3

Q = AV
=
(0.07
m)(2.0
m)
1.24
=
0.174
1 1


s 
s

(1)
Problem 3.126
A Venturi meter with a minimum diameter of 3 in. is to be used to measure the flowrate of
water through a 4-in.-diameter pipe. Determine the pressure difference indicated by the
ft3
and viscous effects are
pressure gage attached to the flowmeter if the flowrate is 0.5
s
negligible.
Solution 3.126
V1
V2
D1 = 4 in.
Q = A2
Water
(2)
(1)
D2 = 3 in.
2( p1 − p2 )
slugs
ft3
, where Q = 0.5
and ρ = 1.94 3
  A 2 
s
ft
ρ 1 −  2  
  A1  


2
A D 
Thus, since 2 =  2  ,
A1  D1 
0.5
ft3 π  3 
=  ft 
s
4  12 
2
2( p1 − p2 )
4
slugs    3 in.  

1.94 3  1 −  4 in.  
ft   

 
or
 slug ⋅ ft 


2
slug
lb
p1 − p2 = 68.8 2 = 68.8  s 2  = 68.8 2
s ft
ft
ft
Problem 3.127
Determine the flowrate through the Venturi meter shown in the figure below if ideal conditions exist.
p1 = 735 kPa
Q
p2 = 550 kPa
19 mm
31 mm
γ = 9.1 kN/m3
Solution 3.127
p1 = 735 kPa
Q
p2 = 550 kPa
31 mm
(1)
(2)
19 mm
γ = 9.1 kN/m3
p1 V12
p2 V22
+
+z =
+
+ z , where z1 = z2 and A1V 1= A 2V 2
γ 2g 1 γ 2g 2
2
D 
A
or V1 = 2 V2 =  2  V2
A1
 D1 
Thus,
4
 D2  2

 V2
p1  D1 
p V2
+
= 2+ 2
γ
2g
γ 2g
or
m  (735 − 550) kPa

2  9.81 2 
(p − p )
s   9.1 kN 

2g 1 2


m
γ
m3 

V2 =
=
= 21.5
4
4
s
D 
 19 mm 
1− 
1−  2 

 31mm 
 D1 
so that
Q = A2V2 =
π
4
D22V2 =
π
m
m3

(0.19 m)2  21.5  = 6.10 ×10 −3
4
s 
s

Problem 3.128
For what flowrate through the Venturi meter shown in the figure below will cavitation
begin if p1 = 275 kPa gage, atmospheric pressure is 101kPa (abs), and the vapor pressure is
3.6 kPa (abs) if ideal conditions exist?
Q
19 mm
31 mm
γ = 9.1 kN/m3
Solution 3.128
Q
31 mm
(1)
(2)
19 mm
γ = 9.1 kN/m3
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
(1)
where z1 = z2 , p2 = 3.6 kPa (abs)
and p1 = (275 + 101)kPa (abs) = 376 kPa (abs)
Thus, with A1V 1= A 2V 2
or
2
D 
V1 =  2  V2 Eq. (1) becomes
 D1 
1
 
m  (376 − 3.6) kPa  2
2  9.81 


(p − p )
kN
s 
2g 1 2
9.1 3
 

γ
m


V2 =
=
4
4


D 
 19 mm 


1− 
1−  2 



 31mm 
 D1 


or
V2 = 30.6
m
s
Thus,
Q = A2V2 =
π
4
D2 2V2 =
π
m
m3

(0.19 m)2  30.6  = 8.68 × 10 −3
4
s 
s

Problem 3.129
What diameter orifice hole, d, is needed if under ideal conditions the flowrate through the
gal
lb
of seawater with p1 − p2 = 2.37 2 ? The
orifice meter of the figure below is to be 30
min
in.
contraction coefficient is assumed to be 0.63 .
p1
p2
d
Q
2-in.
diameter
Solution 3.129
p1
p2
d
(1)
Q
(2)
2-in.
diameter
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 ,
γ 2g
γ 2g
(1)
where z1 = z2 , Cc = 0.63, and p1 − p2 = 2.37 psi
With
3
3 
ft3
lb
 gal   1min   231in.  1ft
Q =  30
0.0668
=
and
64.0
γ
=


 

3
s
 min   60 s   1gal 
ft 3
 1728 in 
it follows that
ft 3
Q
s = 3.06 ft
=
V1 =
s
A1 π  2 2
 ft 
4  12 
0.0668
Thus, Eq. (1) gives
lb 

2.37 × 144 2 

 p − p2 
ft 
ft 


ft 
V2 = V12 + 2 g  1
 =  3.06  + 2  32.2 2  
lb
s

s   64.0

 γ




3
ft


2
or
V2 = 18.8
ft
s
Thus, since
Q = A2V2 = C c
π
4
d 2V2
it follows that
1

1
ft3  2
 4Q  2  4 × 0.0668 s 
 = 0.847ft = 1.016 in.
d =
 =
 π (0.63) 18.8 ft  
 πCcV2 



s  

Problem 3.130
A weir of trapezoidal cross section is used to measure the flowrate in a channel as shown in
the figure below. If the flowrate is Q0 when H =
H
2
, what flowrate is expected when H = ?
30°
ℓ
Solution 3.130
Q = AV where it is expected that V is a function of the head, H.
That is, V
2 gH
1
Also, from the geometry A = H ( + T ) where
2
T =
+ 2H tan30°
Thus, A = H ( + H tan30°) so that
Q = C1 2 g (
3
+ H tan30°)H 2
where C1 is a constant
Let Q0 = flowrate when H =
2
and Q = flowrate when H =
Thus,
3

  2
 1

1 + tan30° 
C1 2 g  + tan30°   

Q0
2

 2  =  2
 = 0.289
=
3
Q
 3
C1 2 g ( + tan30°)( ) 2
(1 + tan30°)  2 2 
 
 
or
Q = 3.46Q0
Problem 3.131
The flowrate in a water channel is sometimes determined by use of a device called a Venturi
flume. As shown in the figure below, this device consists simply of a bump on the bottom of
the channel. If the water surface dips a distance of 0.07 m for the conditions shown, what
is the flowrate per width of the channel? Assume the velocity is uniform and viscous effects
are negligible.
0.07 m
V1
V2
1.2 m
0.2 m
Solution 3.131
0.07 m
(1)
(2)
V1
V2
1.2 m
0.2 m
p1 V12
p2 V22
+
+z =
+
+ z with p1 = 0,
γ 2g 1 γ 2g 2
p2 = 0, z1 = 1.2 m,
and z2 = 1.2 m − 0.07 m = 1.13 m
Also, AV
1 1 = A2V2
or
V2 =
h1
1.2 m ×V1
V1 =
= 1.29V1
h2
(1.2 − 0.07 − 0.2) m
Thus, from Eq. (1):
V12
V22
+ z1 =
+ z2
2g
2g
m

or [(1.29)2 − 1]V12 = 2  9.81 2  (1.2 − 1.13) m
s 

or V1 = 1.438
m
s
Hence,
m
m2

g = hV
1 1 = 1.438
 (1.2 m) = 1.73
s 
s

(1)
Problem 3.132
Water flows under the inclined sluice gate shown in the figure below. Determine the
flowrate if the gate is 8ft wide.
30°
6 ft
1.6 ft
1 ft
Solution 3.132
(1)
30°
6 ft
(2)
1.6 ft
1 ft
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p1= 0, p 2 = 0, z 1= 6 ft and z2 = 1ft
γ 2g
γ 2g
Thus,
V12
V2
+ 6 ft = 2 + 1ft
2g
2g
(1)
But A1V 1= A 2V 2, or
V2 =
A1
6 ft
V1 =
V1 = 6V1
1ft
A2
Hence, Eq. (1) becomes
(6)2V12
V12
+ 6 ft =
+ 1ft
2g
2g
or
ft 

[6 2 − 1]V12 = 2  32.2 2  (6 − 1) ft
s 

or
Hence,
ft 
ft3

Q = AV
=
6
ft
(8ft)
3.03
=
145
1 1


s
s

V1 = 3.03
ft
s
Problem 3.133
Water flows in a vertical pipe of 0.15 m diameter at a rate of 0.2 m3 / s and a pressure of
200 kPa at an elevation of 25m . Determine the velocity head and pressure head at
elevations of 20 and 55m .
Solution 3.133
V=
m3
Q = 0.2 s
m3
0.2
s
m
Q
=
= 11.3 = V0 = V2
π
A
s
(0.15 m)2
4
z2 = 55 m
At point (0):
2
m

11.3 
V02 
s 
=
= 6.51m
m
2g

2  9.81 2 
s 

D = 0.15 m
z1 = 25 m
and
V02
p1 V12
+
+z = +
+ z or
γ 2g 0 γ 2g 1
p0
γ
200
=
p0
γ
=
p1
γ
z0 = 20 m
+ z1 − z0
kN
m 2 + (25 − 20) m = 25.4 m
kN
9.8 3
m
Similarly, at point (2):
V02 V22
=
= 6.51m
2g 2g
and
p2
γ
or
+
(1)
p1 = 200 kPa
p0
or
(2)
V22
p V2
+ z2 = 1 + 1 + z1 or
2g
γ 2g
p2
γ
200
=
p2
γ
=
p1
γ
+ z1 − z2
kN
m 2 + (25 − 55) m = −9.59 m .
kN
9.8 3
m
(0)
Problem 3.134
Water flows from a large tank as shown in the figure below. Atmospheric pressure is
14.5 psia . Draw the energy line and the hydraulic grade line for the flow.
(0)
h
D3 = 4 in.
(2)
D1 = 1 in.
D2 = 2 in.
Solution 3.134
For inviscid flow with no pumps or turbines, the energy line is horizontal, a distance h
above the outlet. From Problem 3.90, we obtain h = 1.79ft .
The hydraulic grade line is
V2
below the energy line, starting at the free surface where
2g
V0 = 0 and ending at the pipe exit where p2 = 0 and
V22
= h . At point (1), the pressure
2g
lb  144 in.2 
(2.88 − 14.5) 2 

in.  ft 2 
p1
=
= −26.8ft , and z1 = 0
head is
lb
γ
62.4 3
ft
2
D 
AV
In the 4-in. pipe, V3 = 2 2 =  2  V2 so that
A3
 D3 
4
4
4
V32  D2  V2 2  D2 
2
=
=

 h =   (1.79 ft) = 0.112 ft
2 g  D3  2 g  D3 
4
The corresponding EL and HGL are drawn to scale below.
V32
= 0.112 ft
2g
(0)
(3)
(1)
V22
= h = 1.79 ft
2g
pipe center line
Hydraulic Grade
Line (HGL)
Energy Line (EL)
(2)
z=0
p1
γ = –26.8 ft
Problem 3.135
Water flows steadily with negligible viscous effects through the pipe shown in the figure below. Draw the energy line and hydraulic grade line for the flow.
(1)
4 ft
4-in.-diameter thin-walled tubing
(2)
h
6 in.
(3)
Solution 3.135
For steady, inviscid flow with no pumps or turbines the energy line is horizontal, a distance
of h + 4 ft = 1.31ft + 4 ft = 5.31ft above the outlet. (See solution to Problem 3.77 for values
of h, p2 ,V2 , and p3 ,V3 ).
V2
, below the energy line.
The hydraulic grade line is one velocity head,
2g
2
ft 

41.7 
2

2
V
V
s
Thus, with 1 = 0 2 = 
= 27.0 ft ,
2g
2g  
ft  
 2  32.2 2  
s 
 
and
2
ft 

18.5 
2

V3
s
= 
= 5.31ft
2g  
ft  
 2  32.2 2  
s 
 
The following EL and HGL are obtained:
Energy Line (EL)
(1)
z2 = 0
(2)
V2
5.31 ft =
z3 = –1.31 ft
V32
2g
(3)
V3
V22
= 27.0 ft
2g
Hydraulic Grade Line (HGL)
p2
γ = –23 ft
lb
p
ft 2 = 23ft
Note: 2 = −
lb 
γ

 62.4 3 
ft 

144
Problem 3.137
The following table lists typical flight speeds for two aircraft. For which of these conditions
would it be reasonable to use the incompressible Bernoulli’s equation to study the aerodynamics associated with their flight? Explain.
Flight speed (km/hr)
Aircraft
Cruise
Landing approach
Boeing 787
F-22 fighter
913
1960
214
250
Solution 3.137
Assume incompressible equations are valid if Ma =
V
< 0.3 .
c
For standard sea level conditions, T = 15°C so that
1
 
2
N⋅m 
c = kRT = 1.4  2.87 × 102
 (15 + 273) K 
kg ⋅ K 
 

m 1km 3600 s
km
= 340 ×
×
= 1220
s 1000 m 1hr
hr
Thus, for Boeing 787 at cruise speed,
km
hr = 0.748 > 0.3
Ma =
km
1220
hr
913
so that it is not reasonable to use incompressible analysis.
The table below gives results for the four conditions given above.
Aircraft
Cruise Ma
Landing approach Ma
787
0.748 compressible
0.175 incompressible
F-22
1.61 compressible
0.205 incompressible
Note: If the 787 cruises at an altitude of 10,000 m where the temperature is T = −49.9°C
(see Table C.2 Properties of the U.S. Standard Atmosphere [SI Units]), then
m
km
913
c = 300 = 1080
= 0.845
and Ma =
s
hr
1080
Problem 3.138
A meteorologist uses a Pitot-static tube to measure the wind speed in a tornado. Based on
the damage caused by the storm, the tornado is rated as EF5 on the Enhanced Fujita Scale.
This means that the wind speed is estimated to be in the range of 261 to 318 mph. Is it reasonable to use the incompressible Pitot-tube equation [Eq. (3.16)] to determine the actual
wind speed, or must compressible effects to taken into account? Explain.
Solution 3.138
V = 261
miles 
fps 
ft
1.47
 = 383.7
hr 
mph 
s
V = 318
miles 
fps 
ft
1.47
 = 467.5
hr 
mph 
s
For air at standard conditions, c ≈ 1117
ft
(see Table B.3 Physical Properties of Air at
s
Standard Atmospheric Pressure)
ft
ft
487.5
s = 0.344 and Ma =
s = 0.418
Thus, Ma =
ft
ft
1117
1117
s
s
383.7
Since these are both Ma > 0.3 , the flow should not be considered incompressible.
Problem 4.2
The surface velocity of a river is measured at several locations x and can be reasonably
represented by
V = V0 + Δ V (1 − e − ax ) ,
where V0 , ΔV , and a are constants. Find the Lagrangian description of the velocity of a
fluid particle flowing along the surface if x = 0 at time t = 0.
Solution 4.2
GIVEN: See sketch.
V = V0 + ΔV (1–e–ax)
FIND: V (V0 , t ) since we want the velocity of a fluid
particle (Lagrangian description).
SOLUTION: The position x is found from
x
dx
= V = V0 + ΔV (1 − e − ax ).
dt
V = V0
at
x=0
at
t=0
Separating the variables and integrating gives
x
dx
 (V + ΔV ) − ΔVe−ax )
0
0
t
=  dt
0
or
x


x
1
ln(V0 + ΔV − ΔVe − ax )  = t
−

V0 + ΔV (V0 + ΔV )( −a )
0
Substituting the limits gives
 V + ΔV − ΔVe −ax 
x
1
+
ln  0
 = t
V0 + ΔV a(V0 + ΔV ) 
V0

or
1  ΔV (1 − e − ax ) 
x + ln 1 +
 = (V0 + ΔV )t .
a 
V0

Using the original equation for V gives
V − V0
V − V0
,
= 1 − e − ax , e − ax = 1 −
ΔV
ΔV
and
1  V −V0 
x = − ln 1 −
.
a 
ΔV 
(1)
Substituting these expressions for e − ax and x into Eq. (1) gives
1  V −V0  1  ΔV +V −V0 
− ln 1 −
+ ln 
 = (V0 + ΔV )t,
ΔV  a 
a 
V0


1 V
ΔV
ln 
 = (V0 + ΔV )t ,
a  V0 ΔV −V +V0 
V
ΔV
= ea (V0 +ΔV )t ,
V0 ΔV −V +V0
V Δ V = V0 (V0 + ΔV − V )e a (V0 + ΔV )t ,
V ( Δ V + V0 e a (V0 + ΔV )t ) = V0 (V0 + Δ V )e a (V0 + ΔV )t
or
V =
V0 (V0 + ΔV )
V0 + ΔV0 e − a (V0 +ΔV )t
Problem 4.3
m
The velocity field of a flow is given by V = 2 x 2t+ [4 y(t − 1) + 2 x 2t ] where x and y are in
s
meters and t is in seconds. For fluid particles on the x axis, determine the speed and
direction of flow.
Solution 4.3
y
V
θ
θ
x
u = 2x2t, v = 4 y(t − 1) + 2 x 2t , w = 0
For particles on the x-axis y = z = 0, so that
u = 2x2t and v = 2x2t
Thus, on the x-axis, the velocity is in the x − y plane with
V = u2 + v2 + w2 = (2x2t )2 + (2x2t )2
or
V = 2 2 x 2t
m
s
and
tan θ =
v 2 x 2t
=
=1 or θ = 45°
u 2 x 2t
Problem 4.4
A two-dimensional velocity field is given by u = 1 + y and v = 1. Determine the equation of
the streamline that passes through the origin. On a graph, plot this streamline.
Solution 4.4
u = 1 + y and v = 1 so the streamlines are given by
dy v
1
= =
dx u 1 + y
Thus
 (1 + y)dy = dx or
1
y + y2 = x + c, where c is a constant.
2
For the streamline that goes through x = y = 0, c = 0.
Hence,
x = y+
1 2
y
2
This streamline is plotted below. Note that since v = 1 > 0, the direction of flow is as
shown.
3
2
y
1
0
–2
0
2
–1
–2
–3
x
4
6
Problem 4.5
Streamlines are given in Cartesian coordinates by the equation.

y 
ψ = U  y − 2 2 . x 2 + y 2 ≥ 1.
x +y 

Plot the streamlines for ψ = 0, ± 0.0625, and describe the physical situation represented by
this equation. The parameter U is an upstream uniform velocity of 1.0 ft/s.
Solution 4.5
GIVEN: Streamline given by

y 
ψ = U  y − 2 2 . x 2 + y 2 ≥ 1.
x +y 

With U an upstream uniform velocity with ψ = 0 and ±0.0625. U = 1.0 ft/s.
PLOT: Streamlines and describe physical situation.
SOLUTION:
For ψ = 0,
0 = y−
y
2
x + y2
gives
y = 0 and x 2 + y 2 = 1:
The first is a line alone the x-axis and has little meaning in this problem (experience!).
The second is the equation of circle,
x 2 + y 2 = 1.
For ψ = A( = ±0.0625),
y
A= y− 2
x + y2
( y − A)( x 2 + y 2 ) = y,
( y − A)x 2 = ( A − y ) y 2 + y,
or
x=±
y3 − Ay2 − y
A− y
We now tabulate x versus y, for A= 0.0625 and A= −0.0625
A= −0.0625
A= 0.0625
y
x
y
x
0
0.125
0.250
0
±1.409
±1.127
0
0.125
0.250
0
±0.807
±0.859
0.375
0.500
0.625
0.750
0.875
1.000
±1.029
±0.945
±0.849
±0.727
±0.558
±0.258
0.375
0.500
0.625
0.750
0.875
1.000
y
x
y
x
0
−0.125
−0.250
−0.375
−0.500
−0.625
−0.750
−0.875
−1.000
0
±0.807
±0.859
±0.846
±0.799
±0.720
±0.600
±0.410
0
−0.125
−0.250
−0.375
−0.500
−0.625
−0.750
−0.875
−1.000
0
±1.409
±1.127
±1.029
±0.945
±0.849
±0.727
±0.558
±0.258
−
±0.846
±0.799
±0.720
±0.600
±0.410
−
The values in the upper right section and lower left section of the table are not relevant
since they do not satisfy the condition that x 2 + y 2 ≥ 1. The x , y values on the upper left
section and the lower right section are plotted below along with the circle for ψ = 0. The
figure represented the flow over an infinitely long cylinder perpendicular to the paper.
y
1.0
ψ = +0.0625
ψ =0
x
–2.0
–1.0
2.0
1.0
ψ = –0.0625
–1.0
Problem 4.6
A flow can be visualized by plotting the velocity field as velocity vectors at representative
locations in the flow as shown in the figure below.
2V0
y
2ℓ
V0
V
ℓ
θ
v
u
(b)
V0/2
0
2V0
V0
x
y
−ℓ
V0
V=0
x
−2ℓ
2V0
(c)
(a)
10
10
, and vθ = . This flow
r
r
approximates a fluid swirling into a sink as shown in the figure below.
Consider the velocity field given in polar coordinates by vr = −
vθ
vr
r
θ
Plot the velocity field at locations given by r = 1, 2, and 3 with θ = 0 , 30 , 60 , and 90.
Solution 4.6
With Vr = −
10
10
and Vθ =
then
r
r
2
V=
Vr2 + Vθ2 =
V
α
2
14.14
 10   10 
−  +  =
r
 r   r 
– vr
θ
The angle α between the radial direction and the velocity vector
is given by
10
Vθ
r
tan α =
=
=1
 10 
−Vr
− − 
 r 
Thus, α = 45 ° for any r, θ
0
(i.e., the velocity vector is always oriented 45 ° relative to radial lines)
Note: V is independent of θ .
θ = 60
3
2
1
θ = 30
45°
V = 7.07 at r = 2
0
V = 14.14 at r = 1
vθ
V = 4.71 at r = 3
θ=0
Problem 4.7
A car accelerates from rest to a final constant velocity Vf and a police officer records the
following velocities at various locations x along the highway:
x = 0.
x = 100ft
x = 200ft
x = 300ft
x = 400ft
x = 1000ft
V = 0 mph
V = 34.8mph
V = 47.6 mph
V = 52.3mph
V = 54.0 mph
V = 55.0 mph = Vf
Find a mathematical Eulerian expression for the velocity V traveled by the car as a function
of the final velocity Vf of the car and x if t = 0 at V = 0. [Hint: Try an exponential fit to the
data.]
Solution 4.7
GIVEN: Car accelerates from rest to Vf with
V=0
x=0
= 100ft
= 200ft
= 300ft
= 400ft
= 1000ft
= 34.8mph
= 47.6 mph
= 52.3mph
= 54.0 mph
= 55.0 mph = V f
FIND: A mathematical expression for V = (V f , t ) if V = 0 at t = 0
SOLUTION: At suggested in the problem statement, we try an exponential form
V = A + Be kx
where A, B, k are constants. Choosing three corresponding values of V and x gives
V1 = A + Be kx1
(1)
V2 = A + Be kx2 (2)
V3 = A + Be kx3 (3)
Subtracting Eq. (1) from (2) and (2) from (3) gives
V2 −V1 = B ( e kx2 − e kx1 )
(4)
V3 − V2 = B ( e kx3 − e kx2 )
(5)
Dividing Eq. (4) by (5) gives
V2 − V1 e kx2 − e kx1
=
.
V3 − V2 e kx3 − e kx2
Choosing x1 = 0, x2 = 200, and x3 = 1000 gives
47.6 − 0
e 200 k − 1
= 1000 k
55.0 − 47.6 e
− e 200 k
or
6.43 =
e 200 k − 1
e1000 k − e 200 k
Solving for k by trial and error gives k = −0.010025
Equation (4) then gives
B=
e
V2 −V1
=
kx2
kx1
−e
47.6 − 0
e
−0.010025(200)
−1
= −55.0.
Equation (1) gives
A = V1 − Be kx1 = 0 + 55.01(1) = 55.0.
Then
V = 55.0(1 − e −0.010025 x ) or
V = V f (1 − e −0.010025x )
Note: One could easily use a spreadsheet or curve-fitting to find this result.
Problem 4.8
The components of a velocity field are given by u = x + y , v = xy3 + 16 , and w = 0 .
Determine the location of any stagnation points (V = 0) in the flow field.
Solution 4.8
V = u 2 + v 2 + w 2 = ( x + y )2 + ( xy3 + 16)2 = 0
or
u = x + y = 0 so that x = − y
and
v = xy3 + 16 = 0 so that xy3 = −16
Hence, ( − y ) y3 = −16 or y = 2
Therefore, V = 0 at x = −2, y = 2
Problem 4.9
A two-dimensional, unsteady velocity field is given by
u = 5x(1 + t ) and v = 5 y( −1 + t ) ,
where u is the x -velocity component and v the y -velocity component. Find x(t ) and
y(t ) if x = x0 and y = y0 at t = 0 . Do the velocity components represent an Eulerian
description or a Lagrangian description?
Solution 4.9
GIVEN: Two-dimensional, unsteady flow with
u = 5x(1 + t ) =
dx
dt
v = 5 y( −1 + t ) =
FIND: x(t ) and y(t ) if x = x0 and y = y0 at t = 0 .
SOLUTION: Start with
dx
= 5x(1 + t )
dt
dy
= 5 y( −1 + t ) .
dt
Separating variables and integrating gives
x
t
y
t
dx
 x =  5(1 + t )dt ,
0
x
dy
 y =  5( −1 + t )dt ,
0
y
 x  5
ln   = [(1 + t )2 ]t0,
 x0  2
 y  5
ln   = [( −1 + t )2 ]t0
 y0  2
5
= [(1 + t )2 − 1]
2
5
= [( −1 + t )2 − 1]
2
0
+
0
− +
and
=
=
The velocity components is represented an
Eulerian description.
dy
dt
Problem 4.10
The velocity field of a flow is given by u =
−V0 y
1
2
2 2
(x + y )
V0 x
and v =
(x
2
1
2 2
+y )
, where V0 is a
constant. Where in the flow field is the speed equal to V0 ? Determine the equation of the
streamlines and discuss the various characteristics of this flow.
Solution 4.10
y
u = −V0
1
2
2 2
(x + y )
, v = V0
x
1
2
2 2
(x + y )
so that
1
V02 ( y2 + x2 )  2
2
2
V = u +v = 
 = V0
2
2
 x + y

Thus, V = V0 throughout the entire flow field
Streamlines are given by
dy v x
= =
or − ydy = xdx which can be integrated to give
dx u − y
+
=
Thus, the fluid flow with circular streamlines and the speed is constant throughout.
Problem 4.11
A velocity field is given by
=
+
−
+
, where u and v are in ft/s and x and y
are in feet. Plot the streamline that passes through x = 0 and y = 0 . Compare this
streamline with the streakline through the origin.
Solution 4.11
u = x , v = x( x − 1)( y + 1) where the streamlines are obtained from
dy v x( x − 1)( y + 1)
= =
= ( x − 1)( y + 1)
dx u
x
or 
dy
= ( x − 1)dx which when integrated gives
( y + 1) 
1
ln( y + 1) = x2 − x + c , where c is a constant
2
(1)
For the streamline that passes through the origin x = y = 0 the value of c is found from Eq.
(1) as
ln(1) = c
or
c=0
1
Thus, ln( y + 1) = x2 − x or
2
=



− 

−
This streamline is plotted below
3.5
3
2.5
2
y
1.5
1
0.5
0
–0.5
–1
0
1
2
3
x
Note: The streamline is symmetrical about its low point of x = 1 , y = −0.393 . At x = y = 0 ,
the velocity is 0. For x < 0 , u < 0 and for x > 0 , u > 0 . Thus, the fluid flows from the origin
( x = y = 0 ).
Since the flow is steady, streaklines are the same as streamlines.
Problem 4.12
From time t = 0 to t = 5 hr radioactive steam is released from a nuclear power plant accident
located at x = −1 mile and y = 3 miles . The following wind conditions are expected:
V = 10− 5mph for 0 < t < 3 hr, V = 15+8mph for 3 < t < 10 hr, and V = 5mph for t > 10 hr.
Draw to scale the expected streakline of the steam for t = 3, 10, and 15 hr .
Solution 4.12
For 0 < t < 3 hr, u = 10 mph and v = −5mph
For 3 < t < 10 hr, u = 15mph and v = 8mph
For t > 10 hr, u = 5 mph and v = 0
The streakline is the location (at time t ) of steam released earlier.
a) At t = 3 hr, steam is still being released. From t = 0 to t < 3 hr, it has traveled in the
dy v
5
direction
= = − = −0.5 and the first of the steam is at
10
dx u
x = −1mi + (10 mph)(3hr) = 29 mi and
y = 3mi + ( −5mph)(3hr) = −12 mi at t = 3hr .
See figure below.
b) At t = 5 hr, steam release stops. From t = 3hr to t = 5 hr, the steams travels
Δx = uΔt = (15 mph)(5 − 3) hr = 30 mi «east» and
Δy = vΔt = (8 mph)(5 − 3) hr = 16 mi «north».
See figure below. For t > 5 hr, the streakline does not «grow» (i.e., no more
steamreleased), it merely maintains its shape it had at t = 5 hr
 du du dv dv

=
=
=
= 0  and translates. From t = 5 hrto t = 10 hr, it moves

 dx dy dx dy

Δx = uΔt = (15 mph)(10 − 5) hr = 75 mi farther «east» and
Δy = vΔt = (8mph)(10 − 5) hr = 40 mi farther «north».
See figure below.
c) For 10 < t < 15hr, the steam moves Δx = (5 mph)(15 − 10) hr = 25 mi «east» and
Δy = vΔt = 0 mi «north».
The above is shown in the figure below.
y, mi
(104, 59) (129, 59)
streakline 60
at time
indicated
(99, 43)
(29, 19)
20
(134, 44)
steam at t = 5 hr
(–1, 3)
(59, 4)
t = 3 hr
20
40
60
80
r
–20
(159, 44)
(74, 43)
40
–20
t = 15 hr
t = 10 hr
(x, y) = (29, –12)
x, mi
100
120
140
160
180
Problem 4.13
The x and y components of a velocity field are given by u = x 2 y and v = − xy 2. Determine the
equation for the streamlines of this flow and compare it with those in the Example below. Is
the flow in this problem the same as that in the Example below? Explain.
GIVEN: Consider the two-dimensional steady flow discussed in Example 4.1,
V = (V0 /  )( − xˆi + yˆj ).
FIND: Determine the streamlines for this flow.
y
SOLUTION:
4
Since
u = ( −V0 /  )x and v = (V0 /  ) y
(1)
it follows that streamlines are given by solution of
the equation
dy v (V0 /  ) y
y
= =
=−
dx u −(V0 /  )x
x
in which variables can be separated and the
equation integrated to give
dy
dx
 y = − x
2
C=4
0
Thus, along the streamline
xy = C , where C is a constant
C=1
2
4
C = –1
C = –4
or
ln y = − ln x + constant
C=9
–2
C = –9
(Ans)
Using different values of the constant C, we can
plot various lines in the x−y plane—the
streamlines. The streamlines for x ≥ 0 are plotted
in Fig. E4.2.A comparison of this figure with Fig.
E4.1a illustrates the fact that streamlines are lines
tangent to the velocity vector.
–4
COMMENT Note that a flow is not completely specified by the shape of the streamlines
alone. For example, the streamlines for the flow with V0/ℓ = 10 have the same shape as
those for the flow with V0/ℓ = −10. However, the direction of the flow is opposite for these
two cases. The arrows in Fig. E4.2 representing the flow direction are correct for V0/ℓ = 10
since, from Eq. (1), u = −10x and v = 10y. That is, the flow is from right to left. For
V0/ℓ = −10, the arrows are reversed. The flow is from left to right.
Solution 4.13
Streamlines are given by
dy
dy
dx
dy v
xy2
y
=−
= = − 2 = − or
which can be integrated as:
dx u
x
y
x
x y
dx
 y = − x
Thus, ln y = − ln y + c, where c is a constant.
Thus, xy = c
x
Note: These streamlines are the same shape (same “flow pattern”) as in the Example given
v
are the same:
in the Problem—but the velocity fields are different. However, the ratios
u
v
xy 2
y
= − 2 = − for this problem
u
x
x y
and
V
( 0 )( + y )
v
y
= l
= − for the Example given in the Problem.
u (V0 )( −x )
x
l
Problem 4.14
In addition to the customary horizontal velocity components of the air in the atmosphere
(the “wind”), there often are vertical air currents (thermals) caused by buoyant effects due
to uneven heating of the air as indicated in the figure below.
y
u0
x
0
y
Assume that the velocity field in a certain region is approximated by u = u0 , v = v0 (1 − )
h
for 0 < y < h , and u = u0 , v = 0 for y > h . Plot the shape of the streamline that passes
through the origin for values of
u0
= 0.5, 1, and 2 .
v0
Solution 4.14
y
u = u0 , v = v0 (1 − ) for 0 < y < h so that streamlines
h
for y < h are given by
 y
v0  1 − 
dy v
h
or 
= = 
dx u
u0

 −

v0
y
Thus, −h ln(1 − ) = x
h
u0



=

Note: The lower limits of integration ( x = 0 , y = 0 ) insure that this equation is for the
streamline through the origin.
This streamline

=− 





 −


 is plotted below.

y/h vs x/h
1
0.8
u0/v0 = 2
0.4
u0/v0 = 1
u0/v0 = 0.5
y/h
0.6
0.2
0
0
0.5
1
1.5
x/h
2
2.5
3
Problem 4.15
A test car is traveling along a level road at 88 km/hr. In order to study the acceleration
characteristics of a newly installed engine, the car accelerates at its maximum possible rate.
The test crew records the following velocities at various locations along the level road:
x=0
x = 0.1 km
x = 0.2 km
x = 0.3 km
x = 0.4 km
x = 0.5 km
x = 1.0 km
km
hr
km
V = 93.1
hr
km
V = 98.3
hr
km
V = 104.0
hr
km
V = 110.3
hr
km
V = 117.2
hr
km
V = 164.5
hr
V = 88.5
A preliminary study shows that these data follow an equation of the form V = A(1 + eBx ) ,
where A and B are positive constants. Find A and B and a Lagrangian expression
V = V (V0 , t ) , where V0 is the car velocity at time t = 0 when x = 0.
Solution 4.15
GIVEN: Car accelerates with
km
V = 88.5
x=0
hr
km
V = 93.1
x = 0.1 km
hr
km
V = 98.3
x = 0.2 km
hr
km
V = 104.0
x = 0.3 km
hr
km
V = 110.3
x = 0.4 km
hr
km
V = 117.2
x = 0.5 km
hr
km
V = 164.5
x = 1.0 km
hr
Data follow equation V = A(1 + e Bx ) .
FIND: Lagrangian expression V = V (V0 , t ) where V0 is car velocity at time t = 0 and x = 0.
SOLUTION: The condition V = V0 at x = 0 gives
V0 = A(1 + e 0 ) = 2 A or A =
V0
.
2
Then
V=
V0
(1 + e Bx )
2
The condition V = 93.1 at x = 0.1 gives
93.1 =
88.5
(1 + e0.1B )
2
or
B=
1  2 × 93.1 
ln 
− 1 = 1.0
0.1  88.5

so
V0
(1 + e x )
2
V=
To find V = V (V0 , t ), we note that
V=
dx V0
=
(1 + e x )
dt
2
or
x
dx
 1+ e
0
t
V0
dt.
2
0
=
x
Carrying out the integration gives
x
 e x  
V0
ln 
=
t,


x 

2
 1 + e   0
 ex 
1 V
− ln = 0 t ,
ln 
 1 + e x 
2 2


 e x  V0
ln 
= t − ln 2,
 1 + e x  2


 1 + e x  −V t
ln  x  = 0 + ln 2 ,
 e 
2


1
ex
+1 = e
 −V0t

+ ln2


2
−V0t
= 2e 2 ,
(1)
1
ex
−V0t
= 2e 2 − 1
ex =
1
−V0t
2e 2 − 1
.
Then


V0 
1
1 + −V t  = V (V0 , t )
V=

0
2
 2e 2 − 1
where V0 = 88.5
(2)
km
.
hr
DISCUSSION: Equation(1) can be written as
t=
2  2e x 
ln 

V0  1 + e x 
which gives the
in the following table for each x .
x (km)
t (s)
V (km/hr)
0
0
0.00110
0.00215
0.00314
0.00407
0.00495
0.00858
88.5
93.1
98.3
104.0
110.3
117.2
164.4
0.1
0.2
0.3
0.4
0.5
1.0
Equation (2) then gives V (km/hr) in the table for each t . These values of V agree with the
given information and we have checked our results.
Problem 4.17
A tornado has the following velocity components in polar coordinates:
Vr = −
C1
C
and Vθ = − 2 .
r
r
Note that the air is spiraling inward. Find an equation for the streamlines. r and θ are
polar coordinates.
Solution 4.17
GIVEN: Tornado with velocity components
C
C
Vr = − 1 and Vθ = − 2 .
r
r
FIND: Equation for the streamline of the air.
SOLUTION: First sketch the flow.
y
Vr
r
Vθ
θ
x
Now
Vr =
dr
dθ
and Vθ = r
.
dt
dt
The streamline is tangent to the velocity vector so
Vr
dr
=
.
Vθ rdθ
Substituting for Vr and Vθ gives
 −C1 


 r  = dr or k = dr = C1
rdθ C2
 −C2  rdθ


 r 
where k is a constant. Separating the variables and integrating gives
1 dr
 dθ = k  r + ln k3
where ln k3 is a constant. Integrating gives
1
1
ln r + ln k3 = ln( rk3k ) .
k
k
Solving for r gives
θ=
rk3k = ekθ or
r = k4ekθ ,
equation of a logarithmic spiral,
where k4 is a constant.
Problem 4.18
THE WIDE WORLD OF FLUIDS
Follow those particles Superimpose two photographs of a bouncing ball taken a short time
apart and draw an arrow between the two images of the ball. This arrow represents an
approximation of the velocity (displacement/time) of the ball. The particle image
velocimeter (PIV) uses this technique to provide the instantaneous velocity field for a given
cross section of a flow. The flow being studied is seeded with numerous micronsized
particles that are small enough to follow the flow yet big enough to reflect enough light to
be captured by the camera. The flow is illuminated with a light sheet from a double-pulsed
laser. A digital camera captures both light pulses on the same image frame, allowing the
movement of the particles to be tracked. Using appropriate computer software to carry out
a pixel-by-pixel interrogation of the double image, it is possible to track the motion of the
particles and determine the two components of velocity in the given cross section of the
flow. Using two cameras in a stereoscopic arrangement, it is possible to determine all three
components of velocity.
Two photographs of four particles in a flow past a sphere are superposed as shown in the
figure below.
y, ft
t=0
t = 0.002 s
–0.4
x, ft
–0.2
a = 0.1 ft
The time interval between the photos is Δt = 0.002 s. The locations of the particles, as
determined from the photos, are shown in the table.
Particle
x at t = 0 s (ft)
x at t = 0.002 s (ft)
1
−0.500
−0.480
2
−0.250
−0.232
3
−0.140
−0.128
4
−0.120
−0.112
(a) Determine the fluid velocity for these particles. (b) Plot a graph to compare the results
 a3 
of part (a) with the theoretical velocity, which is given by V = V0 1 + 3 , where a is the
 x 


sphere radius and V0 is the fluid speed far from the sphere.
Solution 4.18
The fluid velocity (which is assumed to be the same as the particle velocity) can be
estimated by
Δx
V=
Δt
[ −0.480 ft − ( −0.500 ft)]
ft
= 10
Thus, for particle (1): V1 =
0.002 s
s
During to 0.002 s time interval, the average location of the particle was
[( −0.480 ft) + ( −0.500 ft)]
x=
= −0.490 ft
2
By similar calculations, the following experimental results were obtained:
Particle
x (ft)
1
2
3
4
−0.490
−0.241
−0.134
−0.116
V
(ft/s)
10
9
6
4
 a3 
ft
These experimental results and the theoretical results (V = V0 1 + 3 , where V0 = 10 and
 x 
s


a = 0.1ft ) are plotted in the figure below.
10
9
8
7
6
V (ft/s)
5
4
3
2
theory
experiment
–0.5
–0.4
1
–0.3
–0.2
x (ft)
–0.1
0
0
Problem 4.19
Air flows steadily through a circular, constant-diameter duct. The air is perfectly inviscid,
so the velocity profile is flat across each flow area. However, the air density decreases as the
air flows down the duct. Is this a one-, two-, or three-dimensional flow?
Solution 4.19
GIVEN: In viscid, steady, flow of air through a circular, constant diameter, duct with a flat
velocity profile at each flow area. Air density decreases as the air flows down the duct.
FIND: Directionality and dimensionality of flow.
SOLUTION: Since the velocity vector has only one component (in the duct’s axial
direction), the flow is
one-dimensional
Problem 4.20
A constant-density fluid flows in the converging, two-dimensional channel as shown in the
figure below. The width perpendicular to the paper is quite large compared to the channel
height. The velocity in the z direction is zero. The channel half-height, Y , and the fluid x
velocity, u, are given by
Y=
2
Yo
 x   y  
and u = uo 1 +  1 −    ,
1+ x 
     Y  
where x, y, Y , and  are in meters , u is in m/s, uo = 1.0 m/s, and Yo = 1.0 m. (a) Is this
flow steady or unsteady? Is it one-dimensional, two-dimensional, or three-dimensional?
(b) Plot the velocity distribution u ( y ) at x /  = 0, 0.5, and 1.0. Use y /Y values of 0,
±0.2, ±0.4, ±0.6, ±0.8, and ±1.0.
y
Y0 = 1.0 m
Y(x)
u(x, y)
+
Yℓ = 0.5 m
x
ℓ = 5.0 m
Solution 4.20
GIVEN: Channel with x-velocity distribution.
FIND: Is the flow steady or unsteady, one-, two-, or three-dimensional. Plot velocity
distribution.
SOLUTION: Because no time variation is given, the flow must be steady.
An x-velocity is given. Because the channel is converging there must be a y-direction
velocity toward the centerline. Because the channel is very wide perpendicular to the paper,
there is no z-velocity. The flow is two-dimensional.
ANSWER: Substitute numerical values of x /  and y / Y into equation for u ( x , y ) . The
results are given in Table 4.20 and plotted in Fig. 4.20.
Table 4.20. Tabulated values of the velocity u ( x, y ) as a function of the length ratio
the height ratio y /Y , and the height y for the two-dimensional channel.
Length Ratio x/ℓ
Height Ratio y/Y
Height y (m)
Fluid Velocity u (m/s)
0.0
0.00
±0.20
±0.40
±0.60
±0.80
±1.00
0.00
±0.20
±0.40
±0.60
±0.80
±1.00
1.00
0.96
0.84
0.64
0.36
0.00
0.5
0.00
±0.20
±0.40
±0.60
±0.80
±1.00
0.00
±0.13
±0.27
±0.40
±0.53
±0.67
1.50
1.44
1.26
0.95
0.54
0.00
1.0
0.00
±0.20
±0.40
±0.60
±0.80
±1.00
0.00
±0.10
±0.20
±0.30
±0.40
±0.50
2.00
1.92
1.68
1.28
0.72
0.00
Figure 4.20. Plot of the velocity u ( x, y ) as a function of the transverse coordinate y at
various axial locations.
y (m)
y
1
1
1
u (m/s)
0
u
0
1
–1
y
–1
u
0
1
1
Channel
wall
–1
2
,
Problem 4.22
Classify the following flows as one-, two-, or three-dimensional. Sketch a few streamlines
for each.
(a) Rainwater flow down a wide driveway
(b) Flow in a straight horizontal pipe
(c) Flow in a straight pipe inclined upward at a
angle
(d) Flow in a long pipe that follows the ground in hilly country
(e) Flow over an airplane
(f) Wind blowing past a tall telephone
(g) Flow in the impeller of a centrifugal pump
Solution 4.22
(a) One-dimensional
(b) One-dimensional
(c) One-dimensional
(d) One-dimensional
(e) Three-dimensional
(f) Two-dimensional
(g) Three-dimensional
Problem 4.23
The velocity components of u and v of a two-dimensional flow are given by
bx
by
u = ax + 2 2 and v = ay + 2 2 ,
x y
x y
where a and b are constants. Calculate the acceleration.
Solution 4.23
GIVEN: Velocity components
by
and
,
v
ay
=
+
x2 y2
x2 y2
where a and b are constants.
u = ax +
bx
FIND: Accelerations ax and a y .
SOLUTION: Since the flow is spatially two-dimensional ( x, y ) and steady, the equations
are
ax = u
∂u
∂u
∂v
∂v
+v
and ay = u + v .
∂x
∂y
∂x
∂y
Substituting into the equation for ax gives

b  ∂ 
b  
b  ∂ 
b 
ax =  ax + 2   ax + 2  +  ay + 2   ax + 2  =
xy  ∂x 
xy  
x y  ∂y 
xy 


b 
b  
b   2b 
a
ay
=  ax + 2 
−
+
+


2 2 
2 

  − xy3  =
xy
x
y
x
y


 


= a2 x +
= a2x −
ab
ab
xy
xy
−
2
2ab
xy
−
2
−
2
b2
2ab 2b2
−
−
=
x3 y 4 xy2 x3 y 4
 2 2ab
3b2 
x
a
=
−
−


x3 y 4
x 2 y2 x 4 y 4 

3b2
or

3b 
b 
ax = x  a − 2 2 
a
+

x y 
x2 y2 


Substituting into the equation for a y gives

b  ∂ 
b  
b  ∂ 
b 
ay =  ax + 2   ay + 2  +  ay + 2   ay + 2  =

xy  ∂x 
x y  
x y  ∂y 
x y 


b  2b  
b 
b 
=  ax + 2 
−
+
ay
+
a
−



=
3  
xy 
x 2 y 
x 2 y2 

 x y  

=−
2ab
x2 y
= a2 y −
−
2b 2
ab
ab
b2
2
+
a
y
+
−
−
=
x 4 y3
x 2 y x 2 y x 4 y3
2ab
x2 y
−
 2 2ab
3b2 
−
=
y
a
−



x 2 y2 x 4 y 4 
x 4 y3

3b2
or

3b 
b 
ay = y  a − 2 2 
a
+

x y 
x2 y2 


Problem 4.24
Air is delivered through a constant-diameter duct by a fan. The air is inviscid, so the fluid
velocity profile is “flat” across each cross section. During the fan start-up, the following
velocities were measured at the time t and axial positions x :
t=0 s
t = 1.0 s
t = 2.0 s
t = 3.0 s
x=0
m
V =0
s
m
V = 1.00
s
m
V = 1.70
s
m
V = 2.10
s
x = 10 m
m
V =0
s
m
V = 1.20
s
m
V = 1.80
s
m
V = 2.15
s
x = 20 m
m
V =0
s
m
V = 1.40
s
m
V = 1.90
s
m
V = 2.20
s
Calculate the local acceleration, the convective acceleration, and the total acceleration at
t = 1.0 s and x = 10 m . What is the local acceleration after the fan has reached a steady air
flowrate?
Solution 4.24
GIVEN: Air velocity in duct, axial position x at time t .
t=0 s
= 1.0 s
= 2.0 s
= 3.0 s
x=0
m
V =0
s
m
= 1.00
s
m
= 1.70
s
m
= 2.10
s
x = 10 m
m
V =0
s
m
= 1.20
s
m
= 1.80
s
m
= 2.15
s
x = 20 m
m
V =0
s
m
= 1.40
s
m
= 1.90
s
m
= 2.20
s
∂V
∂V
and convective acceleration V
at t = 1.0s and x = 10 m.
∂t
∂x
SOLUTION: The local acceleration is
FIND: Local acceleration
m
m

1.80 − 0 
∂V 
s
s
≈
.

∂t  2.0 s − 0 s 

t =1.0s
x =10 m
or
∂V
m
≈ 0.90 2
∂t
s t =1.0s
x =10 m
The convective acceleration is
m
m

1.40 − 1.00 
∂V 
m 
s
s
V
≈ 1.20  
.

∂x 
s   20 m − 0 m 

t =1.0s
x =10 m
V
∂V
m
≈ 0.024 2
∂x
s t =1.0s
x =10 m
When the air flowrate has become steady,
∂V
= 0.
∂t
DISCUSSION: These values are approximations because
evaluated as
ΔV
ΔV
, respectively.
and
Δt
Δx
∂V
∂V
and
have been
∂t
∂x
Problem 4.25
Water flows through a constant diameter pipe with a uniform velocity given by
8  m
V =  + 5  , where t is in seconds. Determine the acceleration at time t =1, 2, and 10 s .
t
 s
Solution 4.25
∂V
a=
+ V ⋅∇V
∂t
8
m
With u = 0, v =  + 5  , w = 0
t
 s
this becomes
 ∂V
∂V  ∂V 
8 m
a=
j = − 2 j 2
+V ⋅
j=
∂y 
∂t
t s
 ∂t
m
Thus, a = −8j 2 at t =1 s
s
m
a = −2.0j 2 at t = 2s
s
and
m
a = −0.08j 2 at t = 10s
s
Problem 4.26
The velocity of air in the diverging pipe as shown in the figure below is given by V1 = 4t
ft
and V2 = 2t , where t is in seconds.
s
V1 = 4t ft/s
V2 = 2t ft/s
(1)
(2)
(a) Determine the local acceleration at points (1) and (2). (b) Is the average convective
acceleration between these two points negative, zero, or positive? Explain.
Solution 4.26
∂u
ft
=4 2
(a)
∂t (1)
s
and
∂u
ft
=2 2
∂t (2)
s
(b) Convective acceleration along the pipe and = u
∂u
∂x
where u > 0. At any time, t , V2 < V1. Thus, between (1) and (2)
Hence, u
∂u V2 −V1
≈
<0

∂x
∂u
< 0 or the average convective acceleration is negative.
∂x
ft
s
Problem 4.27
A certain flow field has the velocity vector
x 2 − y2 ) z
(
y
V=
i+
j+ 2
. Find the acceleration vector for this flow.
2
2
2
2
2
2
2
x
y
+
(x + y ) (x + y )
−2 xyz
Solution 4.27
V = ui + vj + wk
 ∂u
 ∂w
∂u
∂u
∂u   ∂v
∂v
∂v
∂v 
∂w
∂w
∂w 
=
+u
+v
+ w i +  + u
+v
+w  j +
+u
+v
+w
k
∂x
∂y
∂z   ∂t
∂x
∂y
∂z 
∂x
∂y
∂z 
 ∂t
 ∂t
∂u ∂v ∂w
=
=
=0
∂t ∂t ∂t
Using software to perform calculus and algebra results in:
ax =
ay =
az =
(
) − 2xz ( x2 − 3y2 )( x2 − y2 ) − 2xy2
5
5
3
( x 2 + y2 )
( x 2 + y2 )
( x 2 + y2 )
4 xy2 z 2 y2 − 3x 2
(
) + 2 yz 2 ( x2 − y2 )( y2 − 3x2 ) + y ( x2 − y2 )
5
5
3
( x 2 + y2 )
( x 2 + y2 )
( x 2 + y2 )
4 x 2 yz 2 x 2 − 3 y2
(
)z
4
( x 2 + y2 )
x 2 − y2
2
Problem 4.28
Determine the x - component of the acceleration, ax , along the centerline ( y = 0 ) for the flow
of Problem 4.20. Can you determine the acceleration vector at a location not on the
centerline? Why or why not?
y
Y0 = 1.0 m
Y(x)
u(x, y)
+
Yℓ = 0.5 m
x
ℓ = 5.0 m
Solution 4.28
GIVEN: Flow from Problem 4.20
2
Yo
 x   y  
Y=
and u = uo 1 +  1 −    ,
1+ x 
     Y  
FIND: Acceleration along centerline
SOLUTION: As the flow is two-dimensional
 ∂u
∂u   ∂v
∂v 
= u
+ v  ˆi +  u
+ v  ˆj
∂y   ∂x
∂y 
 ∂x
The flow is symmetric about the centerline so v =
Then ax = u
∂v
= 0 at y = 0
∂y
∂u
and ay = 0. Putting y = 0 in the velocity distribution gives the centerline
∂x
velocity
u0 u02
∂u u0
so ax = u0 (1 + x /  ) = (1 + x /  )
=
u = u0 (1 + x /  )(1 − 0 ) . Then
∂x 


 m2 
1 2 
 s   x [ m] 
m  x [ m] 
Substituting numbers ax =
+
=
1
0.2
 1 + 5 m 
5 m 
5 m 
s2 
Problem 4.29
m
s
m
and V2 = 1.0t , where t is in seconds. Determine the local acceleration at points (1) and
s
(2). Is the average convective acceleration between these two points negative, zero, or
positive? Explain.
The velocity of the water in the pipe as shown in the figure below is given by V1 = 0.50t
V1 =
0.50t m/s
V2 =
1.0t m/s
(2)
(1)
Solution 4.29
∂V1
m
= 0.5 2
∂t
s
∂V2
m
= 1.0 2
∂t
s
∂V
> 0 . Also V > 0 so that the
∂x
∂V
convective acceleration, V
, is positive.
∂x
Since V2 > V1 it follows that
Problem 4.30
A shock wave is a very thin layer (thickness = ) in a highspeed (supersonic) gas flow across
which the flow properties (velocity, density, pressure, etc.) change from state (1) to state (2)
as shown in the figure below.
V1
V1
V
V2
V2
ℓ
Shock wave
ℓ
x
If V1 = 1800fps, V2 = 700 fps, and  = 10 −4 in., estimate the average deceleration of the gas as
it flows across the shock wave. How manyg ’s deceleration does this represent?
Solution 4.30
a=
∂V
∂u
+ V ⋅∇V so that with V = u( x )i , a = axi = u i
∂
∂x
Without knowing the actual velocity distribution, u = u( x ), the acceleration can be
approximated as
ax = u
∂u (V1 +V2 ) (V2 −V1 ) (1800 + 700 ) fps ( 700 − 1800 ) fps
≈
=
∂x
2
2

 10 −4 

 ft
12


or
ft
−1.65 ×1011 2
a
ft
s = −5.12 ×109
ax = −1.65 ×1011 2 This is x =
ft
g
s
32.2 2
s
Problem 4.33
As a valve is opened, water flows through the diffuser as shown in the figure below at an
increasing flowrate so that the velocity along the centerline is given by
 x
V = ui = V0 (1 − e − ct )  1 −  i , where u0 , c, and  are constants. Determine the acceleration


ft
as a function of x and t. If V0 = 10 and  = 5ft, what value of c (other than c = 0) is needed
s
to make the acceleration zero for any x at t = 1 s? Explain how the acceleration can be zero
if the flowrate is increasing with time.
y
ℓ/2
u
u = 1– V0(1 – e–ct)
u = V0(1 – e–ct)
2
x
Solution 4.33
∂V
a=
+ V ⋅∇V
∂t
with u = u( x, t ), v = 0, and w = 0
this becomes
∂u 
 ∂u
 x
a =  + u i = axi , where u = V0 (1 − e − ct )  1 − 


∂x 
 ∂t
Thus,
 x
 x   1
ax = V0  1 −  ce − ct + V02 (1 − e − ct )2  1 −   − 



   
or
V

 x 
ax = 1 −  ce −ct − 0 (1 − e −ct )2 
  


If, ax = 0 for any x at t =1 s we must have
 −ct V0
− ct 2 
ce −  (1 − e )  = 0


With V0 = 10 and  = 5
ce −c −
10
(1 − e −c )2 = 0
5
With solution (root) of this equation is c = 0.490
1
s
 ∂u

For the above conditions, the local acceleration  > 0  is precisely balanced by the
 ∂t

∂
u


< 0 . The flowrate increases with the time, but the fluid flows
convective deceleration  u
 ∂x

to an area of lower velocity.
Problem 4.34
m
at point A
s
m
to 18 at point B . It is also known that the velocity is a linear function of distance along the
s
streamline. Determine the acceleration at points A, B , and C. Assume steady flow.
The fluid velocity along the x axis as shown in the figure below changes from 6
VA = 6 m/s
VB = 18 m/s
A
C
B
x
0.05 m
0.1 m
Solution 4.34
∂V
+ V ⋅∇V
∂t
with u = u( x ), v = 0, and w = 0
this becomes
a=
∂u 
∂u
 ∂u
a =  + u i = u i ,
(1)
∂x 
∂x
 ∂t
Since u is a linear function of x, u = c1x + c2 where the constants c1, c2 are given as:
uA = 6 = c2
and
uB = 18 = 0.1c1 + c2
or
c1 = 120, c2 = 6
Thus, u = (120x + 6)
m
with x − m
s
From Eq.(1)
a=u
∂u 
m 
m 
i = (120 x + 6)
120
i
s 
m ⋅s 
∂x
or
m
for xA = 0, a A = 720i 2
s
m
for xB = 0.05 m , a B = 1440i 2
s
m
for xC = 0.1 m, aC = 2160i 2
s
Problem 4.35
x
A fluid flows along the x axis with a velocity given by V =  i , where x is in feet and t in
t
seconds. (a) Plot the speed for 0 ≤ x ≤ 10 ft and t = 3s. (b) Plot the speed for x = 7 ft and
2 ≤ t ≤ 4s. (c) Determine the local and convective acceleration. (d) Show that the
acceleration of any fluid particle in the flow is zero. (e) Explain physically how the velocity
of a particle in this unsteady flow remains constant throughout its motion.
Solution 4.35
x ft
x ft
(a) u =
so at t = 3s, u =
t s
3 s
(b) For x = 7 ft, u =
(c)
7 ft
t s
∂u x  1  x
∂u
x
=  =
= − 2 and u
∂x t  t  t2
∂t
t
(d) For any fluid particle a =
∂V
+ V ⋅∇V which with v = 0 w = 0 becomes
∂t
∂u   x x 
 ∂u
a =  + u i =  − + i ≡ 0
∂x   t2 t2 
 ∂t
(e) The particles flow into areas of higher velocity (see Fig.1), but at any given location, the
velocity is decreasing in time (see Fig.2). For the given velocity field, the local and
convective accelerations are equal and opposite, giving zero acceleration throughout.
4
u, fps
t=3s
3
2
Fig.1
1
0
0
5
10
x, ft
3
u, fps
2
x = 7 ft
1
0
Fig.2
0
1
2
t, s
3
4
Problem 4.36
A constant-density fluid flows through a converging section having an area A given by
A
A= 0
x
1+

whereA0 is the area at x = 0. Determine the velocity and acceleration of the fluid in
Eulerian form and then the velocity and acceleration of a fluid particle in Lagrangian
form. The velocity is V0 at x = 0 when t = 0.
Solution 4.36
GIVEN: Constant-density fluid flowing through converging section with area A0 and
velocity V0 at x = 0 when t = 0. Area given by
A=
A0
.
x
1+

FIND: Velocity and acceleration of fluid in Eulerian form. Velocity and acceleration of
fluid in Lagrangian form.
SOLUTION: Apply the integral continuity equation to the control volume of length x in
the figure.
ℓ
V0
V
x
Since we have a constant density fluid and the control volume is fixed in space, we have
 m in =  m out,
A0V0 = AV ,
 x
A 
V = V0  0 or V = V0 1 +  . Eulerian velocity form
 
 A
The acceleration is found using equation
ax =
∂u
∂u ∂v
∂v
+u
= +v .
∂t
∂x ∂t
∂x
Substituting for V gives
V2  x
 x  V 
ax = 0 + V0 1 +  0  or ax = 0 1 +  . Eulerian acceleration form
  
    
The velocity of a fluid particle is found by first integrating the above equation for V
dx
 x
= V = V0 1 + ,
dt
 
x
t
 x
 ln 1 +  = V0t,
 
0
x
or 1 + = e  .

dx
  x  = V0dt,
0 1 +  0
 
Vt
Substituting this expression into the equation for V gives the particle velocity.
V0t
V = V0 e  . Lagrangian velocity form
Substituting this same expression into the equation for ax gives the particle acceleration.
Vt
V2 0
ax = 0 e  . Lagrangian acceleration form

Problem 4.37
A hydraulic jump is a rather sudden change in depth of a liquid layer as it flows in an open
channel as shown in the figure below.
Hydraulic jump
ℓ
V1
V2
z2
z1
In a relatively short distance (thickness = ) the liquid depth changes from z1 to z 2 , with a
corresponding change in velocity from V1 to V2 . If V1 = 1.20 ft/s , V2 = 0.30ft/s, and  = 0.02 ft,
estimate the average deceleration of the liquid as it flows across the hydraulic jump. How
manyg ’s deceleration does this represent?
Solution 4.37
∂V
∂u
+ V ⋅∇Vso with V = u( x )i , a = axi = u i
a=
∂t
∂x
Without knowing the actual velocity distribution, u = u( x ), the acceleration can be
approximated as
(V −V ) 1
ft
∂u 1
ax = u
≈ (V1 + V2 ) 2 1 = (1.20 + 0.30)
2
s

∂x 2
ft
33.8 2
a
s = 1.05
Thus, x =
ft
g
32.2 2
s
(0.30 − 1.20)
0.02 ft
ft
s = −33.8 ft
s2
Problem 4.38
A fluid particle flowing along a stagnation streamline, as shown in the figure below, slows
down as it approaches the stagnation point. Measurements of the dye flow indicate that the
location of a particle starting on the stagnation streamline a distance s = 0.6 ft upstream of
the stagnation point at t = 0 is given approximately by s = 0.6 e−0.5t , where t is in seconds
and s is in feet. (a) Determine the speed of a fluid particle as a function of time, Vparticle (t ),
as it flows along the streamline. (b) Determine the speed of the fluid as a function of
position along the streamline, V = V ( s ). (c) Determine the fluid acceleration along the
streamline as a function of position, as = as ( s ).
Stagnation point, s = 0
Fluid particle
V
s
Solution 4.38
(a) With s = 0.6 e−0.5t it follows that Vparticle =
ds
ft
= 0.6( −0.5)e −0.5t = −0.3e −0.5t
dt
s
(b) From part (a),
V = ( −0.5) 0.6e −0.5t  where s = 0.6e −0.5t
Thus,
V = ( −0.5)[ s ], or V = −0.5s
ft
where s −ft
s
(c) For steady flow, as = V
dV
ds
Thus, with V = −0.5s and
dV
= −0.5,
ds
ft
as = ( −0.5s )( −0.5) = 0.25s 2 where s −ft
s
Note: For S > 0, as is positive—the particle’s acceleration is to the right. Since the particle is
moving to the left, a positive as for this case implies that the particle is decelerating (as it
must be for this stagnation point flow).
Problem 4.39
A nozzle is designed to accelerate the fluid from V1 to V2 in a linear fashion. That is,
V = ax + b, where a and b are constants. If the flow is constant with V1 = 10 m/sat x1 = 0
and V2 = 25m/sat x2 = 1m , determine the local acceleration, the convective acceleration,
and the acceleration of the fluid at points (1) and (2).
Solution 4.39
With u = ax + b, v = 0, and w = 0 the acceleration a =
∂V
+ V ⋅∇V can be written as
∂t
∂u
a = axi where ax = u .
(1)
∂x
Since u = V1 = 10 m/s at x = 0 and u = V2 = 25 m/s at x =1, we obtain
10 = 0 + b
25 = a + b so that a =15 and b =10
That is, u = (15x + 10) m/s , where x − m , so that from Eq.(1)
ax = (15x + 10)
m  1
m
15  = (225x + 150) 2 .
s  s
s
Note: The local acceleration is zero,
u
∂V
= 0, and the convective acceleration is
∂t
∂u 
m
i = (225x + 150)i 2
∂x
s
m
m
At x = 0, a = 150i 2 ; at x = 1m, a = 375i 2
s
s
Problem 4.40
An incompressible fluid flows through the converging duct as shown in the figure (a) below
with velocity V0 at the entrance. Measurements indicate that the actual velocity of the fluid
near the wall of the duct along streamline A − F is as shown in the figure (b). Sketch the
component of acceleration along this streamline, a , as a function of s . Discuss the
important characteristics of your result.
s
A
B
C
D
E
F
V0
1.5 V0
(a )
V
1.5 V0
V0
A
B
C
D
E
F
(b)
Solution 4.40
∂V
Since as = V
, it follows that as can be obtained from values of V (which is always
∂s
∂V
∂V
which can be positive, zero, or negative. Values of
and the
positive) and the slope
∂s
∂s
corresponding as are as shown in the figures below.
∂V
∂s +
1
s
0
A
B
C
D
E
F
D
E
S
–
as = V ∂ V
∂s +
ft
s2
0
–
A
B
C
F
S
Problem 4.41
ft
Air flows steadily through a variable area pipe with a velocity of V = u( x )i , where the
s
approximate measured values of u( x ) are given in the table. Plot the acceleration as a
function of x for 0 ≤ x ≤ 12 in. Plot the acceleration if the flowrate is increased by a factor
of N (i.e., the values of u are increased by a factor of N ) for N = 2, 4, 10.
0
10.0
10.2
13.0
20.1
28.3
28.4
25.8
1
2
3
4
5
6
7
8
9
10
20.1
17.4
13.5
11.9
10.3
10.0
10.0
11
12
13
Solution 4.41
∂V
+ V ⋅∇V simplifies to a = axi where
∂t
(1)
Since u = u( x ), v = 0, and w = 0, it follows that a =
ax = u
∂u
∂x
∂u
can be obtained by
∂x
an approximate numerical differentiation. The results are shown below for the given date
(i.e., with N =1).
∂u
Note that since ax = u , it follows that and increase in velocity from u to Nu increases the
∂x
2
acceleration from ax to N ax
The values u are given in the table; the corresponding values of
0
1
2
3
4
5
6
7
8
9
10
11
12
13
10
10.2
13
20.1
28.3
28.4
25.8
20.1
17.4
13.5
11.9
10.3
10
10
2.4
18
59.4
91.8
49.8
−15
−49.8
−50.4
−39.6
−33
−19.2
−11.4
−1.8
0
24
184
772
1845
1409
−426
−1285
−1013
−689
−446
−228
−117
−18
0
The results are plotted below.
30
25
u, ft/s
20
15
10
5
0
2
0
4
6
8
10
12
14
x, in.
100
80
du/dx, 1/s
60
40
20
0
0
2
4
6
10
8
12
14
–20
–40
–60
x, ft
2500
N=1
For N ≠ 1 multiply ax
by N2
2000
1500
ax, ft/s2
1000
500
0
–500 0
2
4
6
8
–1000
–1500
x, in.
10
12
14
Problem 4.42
As is indicated in the figure below, the speed of exhaust in a car’s exhaust pipe varies in
time and distance because of the periodic nature of the engine’s operation and the damping
effect with distance from the engine. Assume that the speed is given by
V = V0 1 + ae − bx sin(ωt ) , where V0 = 8fps, a = 0.05, b = 0.2 ft −1, and ω = 50 rad/s. Calculate
and plot the fluid acceleration at x = 0, 1, 2, 3, 4, and 5ft for 0 ≤ t ≤
π
25
s.
5 ft
V
x
V = V0[1 + ae–bx sin(ω t)]
Solution 4.42
Since u = u( x, t ), v = 0, and w = 0 it follows that a =
where ax =
∂u
∂u
+u
∂t
∂x
∂V
+ V ⋅∇V = axi
∂t
(1)
Thus, with u = V0 1 + ae − bx sin(ωt )
Eq.(1) gives
ax = V0 aω e − bx cos(ωt ) + V0 1 + ae − bx sin(ωt ) V0 a( −b )e − bx sin(ωt )
= V0 ae − bx ω cos(ωt ) −V0 b sin(ωt )(1 + ae − bx sin(ωt ))
1
rad
ft
with V0 = 8 , a = 0.05, b = 0.2 , and ω = 50
ft
s
s
this becomes
ax = 0.4e −0.2 x 50 cos(50t ) − 1.6 sin(50t )(1 + 0.05e −0.2 x sin(50t ))
ft
s2
(2)
where t ~ s and x ~ ft
Plot ax from Eq. (2) for 0 ≤ t ≤
π
25
s with x = 0, 1, 2, 3, 4, and 5ft.
An Excel Program was used to calculate ax from Eq. (2). The results are shown below.
Acceleration at various x locations (ft/s2)
x = 0 ft x = 1 ft x = 2 ft x = 3 ft x = 4 ft
20,00
16,37
13,41
10,98
8,99
19,22
15,73
12,88
10,55
8,64
17,24
14,11
11,56
9,46
7,75
14,18
11,61
9,51
7,79
6,38
10,24
8,39
6,87
5,63
4,61
5,67
4,65
3,81
3,12
2,55
0,74
0,61
0,51
0,42
0,34
−4,23
−3,46
−2,83
−2,31
−1,89
−8,93
−7,31
−5,98
−4,90
−4,01
−13,08 −10,71
−8,76
−7,17
−5,87
−16,42 −13,44
−11,00
−9,01
−7,37
−18,73 −15,34
−12,56
−10,28
−8,42
−19,89 −16,29
−13,33
−10,92
−8,94
−19,81 −16,22
−13,28
−10,87
−8,90
−18,51 −15,15
−12,41
−10,16
−8,32
−16,06 −13,14
−10,76
−8,81
−7,21
−12,61 −10,32
−8,45
−6,91
−5,66
−8,37
−6,85
−5,61
−4,59
−3,76
−3,62
−2,96
−2,42
−1,98
−1,62
1,36
1,12
0,92
0,75
0,62
6,26
5,13
4,20
3,44
2,82
10,77
8,82
7,22
5,92
4,84
14,61
11,96
9,80
8,02
6,57
17,54
14,36
11,76
9,63
7,88
19,38
15,87
12,99
10,64
8,71
20,01
16,38
13,41
10,98
8,99
t/s
0,000
0,005
0,010
0,015
0,020
0,025
0,030
0,035
0,040
0,045
0,050
0,055
0,060
0,065
0,070
0,075
0,080
0,085
0,090
0,095
0,100
0,105
0,110
0,115
0,120
0,125
x, ft
Acceleration, ax, vs Time, t
20
0
1
2
3
4
5
15
10
5
ax, ft
s2
0
–5
–10
–15
–20
0,00
x = 5 ft
7,36
7,07
6,34
5,22
3,77
2,09
0,28
−1,55
−3,28
−4,81
−6,04
−6,89
−7,32
−7,29
−6,81
−5,90
−4,63
−3,07
−1,32
0,51
2,31
3,97
5,38
6,45
7,13
7,36
0,02
0,04
0,06
0,08
0,10
0,12
Problem 4.43
Water flows down the face of the dam as shown in the figure below. The face of the dam
consists of two circular arcs with radii of 10 and 20ft as shown. If the speed of the water
1
along streamline A − B is approximately V = (2 gh ) 2 , where the distance h is as indicated,
plot the normal acceleration as a function of distance along the streamline, an = an ( s ).
s
4 ft
h
A
ℛ1 = 10 ft
20°
C
ℛ2 = 20 ft
V
B
Solution 4.43
V2
an =
where R = R1 = 10 ft from A to C, R = R2 = 20 ft from C to B, and V = 2 gh .
R
2 gh
Thus, an =
R
ft 

2  32.2 2  h
s  = 6.44 h ft , where h  ft
(a) From A to C: an = 
(1)
10 ft
s2
From the figure it follows that s = R1θ and h = R1 + 4 − R1 cos θ
Thus, from Eq.(1)
4 ft
h
A
20°
an = 6.44[4 + 10(1 − cos θ )]
ℛ1 = 10 ft
or
ft

 an = (90.2 − 64.4 cos θ ) s 2

 where s = 10 θ ft with θ  rad
ℛ2 = 20 ft
C
θ
D
E
θmax
The above is valid for 0 ≤ θ ≤ θ max , where from the figure 20sin20° = 10 − 10cos θmax , or
θmax = 71.6°
ft 

2  32.2 2  h
s  = 3.22 h ft , where h  ft
(b) From C to B: an = 
20 ft
s2
(2)
From the figure, it follows that h = R2 sin( β + 20°) + 4 and s = sc + R2 β where from part (a)
sc = 10 θmax = 12.5ft
Thus, from Eq.(2)
an = 3.22 ( 20 sin( β + 20° ) + 4 )
A
4 ft
E
and
h
s = 12.5 + 20β
20°
β
C
ℛ2
or
ft

 an = [64.4 sin( β + 20° ) + 12.9] s 2

 where s = 12.5 + 20 β ft with β  rad.
B
This equation, valid for 0 < β ≤ 70° is plotted in the figure below.
ft
For example, if β = 0 (i.e., point C), an = 34.9 2 and s = 12.5ft
s
ft
Note that this value does not agree with an = 69.8 2 at point C calculated in part (a). There
s
is a discontinuity in a n at point C because there is discontinuity in the radius of curvature
ft
from R1 = 10 ft to R2 = 20ft. At point B , β = 70.0° so that an = 77.3 2 and s = 36.9ft
s
The normal acceleration is plotted as a function of distance along the surface.
80
69.8
ft
77.3 2
ft
s
s2
60
an
ft
s2
40
ft
34.9 2
ft
25.8 2
s
s
20
point A, s = 0 point C, s = 12.5 ft
0
0
10
20
s, ft
30
point B, s = 36.9 ft
40
Problem 4.44
Water flows over the crest of a dam with speed V as shown in the figure below. Determine
the speed if the magnitude of the normal acceleration at point (1) is to equal the
acceleration of gravity, g .
(1) V
ℛ = 2 ft
Solution 4.44
an =
ft 
ft
V2
ft

or with an = 32.2 2 , V = an R =  32.2 2  (2 ft) = 8.02 .
s
R
s 
s

Problem 4.45
Water flows under the sluice gate as shown in the figure below. If V1 = 3
normal acceleration at point (1)?
Sluice gate
ℛ = 0.12 m
(1)
V1 = 3 m/s
Solution 4.45
2
 m
3 
2
V1
m
s 
=
= 75 2
an =
R
0.12 m
s
m
, what is the
s
Problem 4.46
A fluid flows past a sphere with an upstream velocity of V0 = 40 m/s as shown in the figure
below. From a more advanced theory, it is found that the speed of the fluid along the front
3
part of the sphere is V = V0 sin θ . Determine the streamwise and normal components of
2
acceleration at point A if the radius of the sphere is a = 0.20 m.
V
40°
V0
A
θ
a
Solution 4.46
3
3 m
m
V = V0 sin θ =  40  sin θ = 60 sin θ
2
2
s 
s
an =
V2
=
R
( 60 sin 40° )
0.2 m
2
2 m
2
s
m
= 7440 2
s
and
∂V
∂V
∂V ∂V ∂θ
, where
= ( 60 sin θ )
=
∂s
∂s
∂s
∂θ ∂s
∂V
From Eq.(1),
= 60 cos θ
∂θ
∂θ
1
=
Also s = aθ = 0.2θ m, where θ  rad, so that
∂s 0.2 m
as = V
Thus, for θ = 40°
m 
m 1
m

as =  60 sin 40°  60 cos 40° 
= 8860 2
s 
s  0.2 m
s

(1)
Problem 4.47
Assume that the streamlines for the wingtip vortices from an airplane (see the figure below)
K
can be approximated by circles of radius rand that the speed is V = , where K is a
r
constant. Determine the streamline acceleration, as, and the normal acceleration, a n, for this
flow.
y
v
u
S
ℛ=r
Solution 4.47
as = V
dV
K dV
, where, since V = ,
=0
ds
r ds
Thus,
as = 0
Also,
2
K 
 
2
V
K2
r
=  = 3
an =
R
r
r
x
Problem 4.48
The velocity components for steady flow through the nozzle as shown in the figure below
  
are = −
are constants. Determine the ratio of
and =  +    , where V0 and
  
the magnitude of the acceleration at point (1) to that at point (2).
y
(1)
(2)
ℓ/2
x
Solution 4.48
(1) a = ax2 + ay2 , where
2
∂u
∂u  V0  V0 
V 
 y
=  − x  −  +V0 1 +  (0) =  0  x
ax = u + v
∂x
∂y     
 
  
and
2
∂v
∂v  V0 
 y  V  V 
ay = u
+v
=  − x  (0) +V0 1 +   0  =  0  (  + y )
∂x
∂y   
       
Thus, from Eq.(1),
V 
a= 0 
  
2
x2 + (  + y )2
so that
V 
a1 =  0 
  
2
V2
2
+ 2 = 0
4

2
V02
0+ =

and
V 
a2 =  0 
  
2
5
4
Hence,
V02 5
a1
5
=  24 =
= 1.118
a2
4
V0

Problem 4.49
Water flows through the curved hose as shown in the figure below with an increasing speed
of V = 10t ft/s , where t is in seconds. For t = 2 s, determine (a) the component of acceleration
along the streamline, (b) the component of acceleration normal to the streamline, and (c)
the net acceleration (magnitude and direction).
V
ℛ = 20 ft
Solution 4.49
∂V
∂V
∂V
(a) as =
, but
+V
=0
∂t
∂s
∂s
Thus,
∂V
ft
as =
= 10 2 for all t .
∂t
s
(b) an =
V2
=
R
ft 2
s2 = 5t 2 ft
20 ft
s2
(10t )2
or
ft
an t =2 s = 5(2)2 =20 2
s
ft
n = 10s + 20
n 2
(c) At t = 2 sa = ass + an 
s
or
1
1
ft
2
2 2
2
2 2
a = [a s + a n ] = [10 + 20 ] = 22.4 2
s
sˆ
n̂
as = 10
θ
an = 20
and
20
θ = arctan __ = 63.4°
10
Problem 4.50
Water flows though the slit at the bottom of a two-dimensional water trough as shown in
the figure below. Throughout most of the trough the flow is approximately radial (along
c
rays from O) with a velocity of V = , where r is the radial coordinate and cis a constant. If
r
m
the velocity is 0.04
when r = 0.1m, determine the acceleration at points A and B .
s
A
V
r
0.8 m
B
0.2 m
O
Solution 4.50
2
V
a = ann + as
s, where an =
= 0 since R = ∞ (i.e., the streamlines are straight)
R
∂V
∂V
c
, where V =
= −V
∂s
∂r
r
m
Since V = 0.04 when r = 0.1m it follows that
s
Also, as = V
2
4 ×10
m

−3 m
orV =
c = Vr =  0.04  ( 0.1m ) = 4 × 10
s 
s
r

Thus,
2
 c  c  c
as = −    − 2  = 3
 r  r  r
At point A:
2
2

−3 m 
×
4
10


s 

−5 m
×
as =
=3.13
10
(0.8 m)3
s2
At point B :
2
2

−3 m 
 4 ×10

s 

−3 m
×
=2.00
10
as =
(0.2 m)3
s2
−3
m
where r  m
s
Problem 4.51
Air flows from a pipe into the region between two parallel circular disks as shown in the
figure below. The fluid velocity in the gap between the disks is closely approximated by
VR
V = 0 , where R is the radius of the disk, ris the radial coordinate, and V0 is the fluid
r
velocity at the edge of the disk. Determine the acceleration for r =1, 2 or 3 ft if V0 = 5 ft/s
and R = 3 ft.
Disks
R
r
Pipe
V0
V
Solution 4.51
2
V
a = ann + as
s, where an =
= 0 since R = ∞ (i.e., the streamlines are straight)
R
VR
∂V
∂V
, where V = 0
=V
∂s
∂r
r
ft
15 ft
, where r ∼ ft
and R = 3 ft, V =
Since V0 = 5
s
r s
Also, as = V
Thus,
2
 ft 
2
2 2
 5  (3ft)
V0 R
225 ft
s
 V0 R  V0 R 

as = 
− 2 =− 3 =−
=
−
, r  ft

r
r3 ft3
r3 s2
 r  r 
ft
At r = 1ft, as = −225 2
s
ft
At r = 2ft, as = −28.1 2
s
ft
At r = 3ft , as = −8.33 2
s
Problem 4.52
Air flows into a pipe from the region between a circular disk and a cone as shown in the
figure below. The fluid velocity in the gap between the disk and the cone is closely
V0 R 2
approximated by V = 2 , where R is the radius of the disk, ris the radial coordinate,
r
and V0 is the fluid velocity at the edge of the disk. Determine the acceleration for
ft
r = 0.5 or 2 ft if V0 = 5 and R = 2 ft.
s
Pipe
Cone
V
Disk
r
R
Solution 4.52
2
V
a = ann + as
s, where an =
= 0 since R = ∞ (i.e., the streamlines are straight)
R
Also, as = V
Thus, V =
∂V
∂V
since r and s are pointed in opposite directions.
= −V
∂s
∂r
V0 R 2
it follows that
r2
2
 ft 
4
2
2
2
2
4
 5  (2 ft)
 V0 R  2V0 R  2V0 R
800 ft
 s
as = −  2  −
=
=
=
, where r  ft

3
5
5
5 2

 r

r
r
r
r
s



At r = 0.5ft, as =
At r = 2ft, as =
800 ft
ft
=
25600
(0.5)5 s2
s2
800 ft
ft
= 25 2
(2.0) s
s
5
2
Problem 4.53
ft
and is being heated, so the fluid
s
°F
temperature T at axial position x increases at a steady rate of 30.0
. In addition, the
min
°F
fluid temperature is increasing in the axial direction at the rate of 2.0 . Find the value of
ft
DT
the material derivative
at position x .
Dt
Fluid flows through a pipe with a velocity of 2
Solution 4.53
GIVEN: Fluid velocity in x direction is 2.0ft/s where fluid temperature is increasing
at a steady rate of 30 °F/min and fluid temperature increases in axial direction at rate
of 2.0 °F/ft.
FIND: Value of Eulerian derivative at axial position x .
SOLUTION: The Eulerian derivative is
dT ∂T
∂T
=
+u
∂t
∂x
dt
where u is the fluid axial velocity.
The numerical values give
dT 30 °F  min   2.0 ft  2.0 °F 
=

+


dt
min  60 s   s  ft 
dT
°F
= 4.5 .
dt
s
Problem 4.54
A gas flows along the x axis with a speed of V = 5x m/s and a pressure of p = 10 x 2 N/m 2,
where x is in meters. (a) Determine the time rate of change of pressure at the fixed location
x =1. (b) Determine the time rate of change of pressure for a fluid particle flowing past x =1.
(c) Explain without using any equations why the answers to parts (a) and (b) are different.
Solution 4.54
(a) Since p = 10 x 2 it follows that
∂p
= 0 for all x .
∂t
(b) With u = 5x, v = 0, w = 0, and p = 10 x 2 it follows that
Dp ∂p
∂p
∂p
∂p
∂p  m 
N 
N
=
+u
+v +w
=u
=  5x  20x 3  = 100x 2
Dt ∂t
∂x
∂y
∂z
∂x 
s 
m 
m s
Thus
Dp
N
= 100 2
Dt x =1m
m s
(c) For this steady flow the pressure at a point is constant (part (a)), but the pressure for a
given particle changes with time (part (b)) because the particle flows into a higher pressure
region.
Problem 4.55
Assume the temperature of the exhaust in an exhaust pipe can be approximated by
T = T0 (1 + ae−bx )[1 + c cos(ωt )], where T0 = 100°C , a = 3, b = 0.03 m −1, c = 0.05, and
ω =100 rad/s. If the exhaust speed is a constant 3 m/s , determine the time rate of change of
temperature of the fluid particles at x = 0 and x = 4 m when t = 0.
Solution 4.55
m
Since u = 3 , v = 0, and w = 0 it follows that
s
DT ∂T
∂T
∂T
∂T
∂T ∂T
∂T
=
+ V ⋅∇T =
+u
+v
+w
=
+u
Dt
∂t
∂t
∂x
∂y
∂z
∂t
∂x
Thus,
DT
= T0 (1 + ae − bx )( −cω sin(ωt )) + uT0 (1 + c cos(ωt ))( −abe − bx )
Dt
when t = 0:
DT
= – abuT0 (1 + c )e-bx , or with the given data,
Dt
DT
1  m 
°C

, where x  m
= −(3)  0.03   3  (100 °C)(1 + 0.05)e −0.03x = 28.4e −0.03x
Dt
m  s 
s

Thus,
DT
°C
= −28.4 at x = 0, t = 0
Dt
s
and
DT
°C
= −25.1 at x = 4 m, t = 0
Dt
s
Problem 4.56
A bicyclist leaves from her home at 9 A.M. and rides to a beach 40 mi away. Because of a
breeze off the ocean, the temperature at the beach remains 60 °F throughout the day. At the
cyclist’s home the temperature increases linearly with time, going from 60 °F at 9 A.M. to
80 °F by 1 P.M. The temperature is assumed to vary linearly as a function of position
between the cyclist’s home and the beach. Determine the rate of change of temperature
observed by the cyclist for the following conditions: (a) as she pedals 10 mph through a
town 10 mi from her home at 10 A.M.; (b) as she eats lunch at a rest stop 30 mi from her
home at noon; (c) as she arrives enthusiastically at the beach at 1 P.M., pedaling 20 mph .
Solution 4.56
From the given data, the temperature, T , varies as a function of location, x , and time, t , as
shown in the figure.
80°
T
65°
60°
75°
(a)
60°
home
t
9
10
11
10
60°
20
(c)
60°
60°
30
beach
40
x, mi
Thus,
65°
(b)
DT ∂T
∂T
=
+u
Dt
∂t
∂x
(a) At x =10 mi and t =10 A.M.,
∂T (75 − 60 ) 15 
=
=
4 hr
4 hr
∂t
and
∂T (60 − 65 )
1 
=
=−
40 mi
8 mi
∂x
12
1
Thus, with u = 10
mi
,
hr

DT 15 
mi  1  
=
+ 10  −
=
2.5

Dt
4 hr
hr  8 mi 
hr
(b) At noon with u = 0 (resting) and
∂T (65 − 60 ) 5 
=
=
4 hr
4 hr
∂t

DT ∂T
∂T ∂T 5 
=
+u
=
=
= 1.25
Dt
∂t
∂x ∂t 4 hr
hr
(b) Upon arrival at the beach with u = 20 mph,

∂T (60 − 80 )
∂T
=
=
−
0.5
= 0, and
∂x
40 mi
mi
∂t
 

∂T
∂T
DT ∂T
mi 
=
+u
=u
= 20  −0.5  = −10
Dt
hr 
mi 
hr
∂t
∂x
∂x
Problem 4.57
Air is delivered through a constant-diameter duct by a fan. The air is inviscid, so the fluid
velocity profile is “flat” across each cross section. During the fan start-up, the following
pressures for the air flow were measured at the time t and axial positions x :
x = 10 m
x = 20 m
x=0
p = 101 kPa
p = 101 kPa
p = 101 kPa
t=0s
t = 1.0 s
p = 121 kPa
p = 116 kPa
p = 111 kPa
t = 2.0 s
p = 141 kPa
p = 131 kPa
p = 121 kPa
t = 3.0 s
p = 171 kPa
p = 151 kPa
p = 131 kPa
Find the local rate of change of pressure
V
∂p
at t = 2.0s and x = 10 m.
∂x
∂p
and the convective rate of change of pressure
∂t
Solution 4.57
GIVEN: Air velocity and pressure in duct.
Axial position x and time t
x=0
t=0s
= 1.0 s
= 2.0 s
= 3.0 s
m
V =0
s
m
= 1.00
s
m
= 1.70
s
m
= 2.10
s
p = 101 kPa
= 121 kPa
= 141 kPa
= 171 kPa
FIND: Local rate of change pressure
t = 2.0s and x = 10 m.
x = 10 m
x = 20 m
m
m
V =0
p = 101 kPa V = 0
p = 101 kPa
s
s
m
m
= 1.40
= 1.20
= 116 kPa
= 111 kPa
s
s
m
m
= 1.90
= 1.80
= 131 kPa
= 121 kPa
s
s
m
m
= 2.20
= 2.15
= 151 kPa
= 131 kPa
s
s
∂p
∂p
and convective rate of change of pressure V
at
∂t
∂x
SOLUTION: The local rate of change of pressure is
∂p  151kPa − 116 kPa 
≈
 t=2.0s
∂t 
3.0 s − 1.0 s
 x=10 m
or
∂p
kPa
≈ 17.5
. t=2.0s
∂t
s
x=10 m
The convective rate of change of pressure is
V
∂p 
m   121kPa − 141kPa 
≈ 1.80  
 t=2.0s
∂x 
s 
20 m − 0 m
 x=10 m
V
∂p
kPa
≈ −1.80
. t=2.0s
∂x
s
x=10 m
DISCUSSION: These values are approximations since
Δp
Δp
and
.
Δt
Δx
∂p
∂p
and
have been evaluated as
∂t
∂x
Problem 4.59
In the region just downstream of a sluice gate, the water may develop a reverse flow region
as is indicated in the figure below. The velocity profile is assumed to consist of two uniform
regions, one with velocity Va =10 fps and the other with Vb = 3 fps. Determine the net
flowrate of water across the portion of the control surface at section (2) if the channel is
20ft wide.
Control surface
Sluice gate
Vb = 3 ft/s
1.8 ft
1.2 ft
(1)
(2)
Va = 10 ft/s
Solution 4.59
ft3
 ft 
 ft 
Q = Va Aa −Vb Ab =  10  (1.2 ft )( 20 ft ) −  3  (1.8ft )( 20 ft ) = 132
s
 s
 s
Problem 4.60
At time t = 0, the valve on an initially empty (perfect vacuum, ρ = 0) tank is opened and air
rushes in. If the tank has a volume of V0 and the density of air within the tank increases as
ρ = ρ∞ (1 − e −bt ) , where b is a constant, determine the time rate of change of mass within the
tank.
Solution 4.60
For t ≥ 0, ρ = ρ0 [1 − e −bt ] so that M = mass of air in tank M = ρV0 = ρ0V0 [1 − e −bt ]
Thus,
dM
= ρ0V0 be − bt
dt
Problem 4.62
Air enters an elbow with a uniform speed of 10 m/s as shown in the figure below. At the exit
of the elbow, the velocity profile is not uniform. In fact, there is a region of separation or
reverse flow. The fixed control volume ABCD coincides with the system at time t = 0. Make
a sketch to indicate (a) the system at time t = 0.01s and (b) the fluid that has entered and
exited the control volume in that time period.
D
1m
V = 10 m/s A
Control volume
1m
5 m/s
Reverse B
flow
C
15 m/s
Solution 4.62
From t = 0 to t = 0.01s particles A, B , C, D, and E move the following distances:


m
 (0.01 s) = 0.1 m = δ D
s 


m
 (0.01 s) = 0.05 m
s 


m
 (0.01 s) = 0.15 m , and
s 
δ A = VAδ t = 10
δ B = VBδ t =  5
δC = VC δ t = 15
δE = 0
Thus, fluid along lines AD and BEC originally moves to lines A′ D ′ and B′E′C′ shown
below.
0.1 m
D
D'
System at t = 0
System at t = 0.015
A'
A
0.1 m
fluid that exited control
volume
B'
0.05 m
E = E'
C
fluid that entered control
volume
B
0.15 m
C'
Problem 4.63
A layer of oil flows down a vertical plate as shown in the figure below with a velocity of
V 
V =  02  (2hx − x2 )j where V0 and h are constants. (a) Show that the fluid sticks to the plate
h 
and that the shear stress at the edge of the layer (x = h ) is zero. (b) Determine the flowrate
across surfaceAB . Assume the width of the plate is b. (Note: The velocity profile for
laminar flow in a pipe has a similar shape.)
x
v(x)
Plate
A
B
Oil
h
y
Solution 4.63
V 
(a) v =  02  (2 hx − x 2 )
h 
Thus,
V 
v x =0 =  02  (0 − 0) = 0 and
h 
τ x =h = μ
V
dv
= μ 02 [2h − 2 x ] x =h = 0
dx x =h
h
Hence, the fluid sticks to the plate and there is no shear stress at the free surface.
x =h
h
(b) QAB =  vdA =  vbdx = 
V0
0h
x =0
2
(2 hx − x 2 )bdx
or
h
V b
1 
2
QAB = 02  hx2 − x3  = V0 hb
3 0 3
h 
Problem 4.64
3
The figure below shows a fixed control volume. It has a volume V0 = 1.0ft , a flow area
A = 1.0 ft 2 , and a length  0 = 1.0 ft . Position x represents the center of the control volume
where the fluid velocity V0 = 1.0ft/s and the density ρ0 = 1.800slug/ft . Also, at position x ,
the fluid density does not change locally with time but decreases in the axial direction at the
dρ
.
linear rate of 0.25 slug/ft 4. Use the system or Lagrangian approach to evaluate
dt
Compare this result with that of the material derivative and flux terms.
3
System and control
volume at time t
ℓ0 = 1 ft
V0, ρ0
A = 1 ft2
+
x0
System at time t + st
Solution 4.64
3
GIVEN: Fixed control volume in the Figure of volume V0 = 1ft . Position x0 (center of
control volume) has velocity V0 = 1ft/s, density ρ0 = 1.800slug/ft , time rate of change of
∂ρ
density
= 0, and spatial linear decreasing rate of change of density = 0.25 slug/ft 4 .
∂t
dρ
FIND:
using Lagrangian or system approach and Eulerian approach.
dt
3
SOLUTION: We will first use the Eulerian approach.
d ρ ∂ρ
∂ρ
 ft   −0.25slug 
=
+v
= 0 + 1 

dt
∂t
∂x
 s 
ft 4

d ρ −0.25slug
=
dt
ft3 s
The density as a function of x is written as
ρ = ρ0 +
dρ
1.800 slug  −0.25slug 
+
( x − x0 ) =
 ( x − x0 )
dx
ft3
ft 4


Now consider a time interval of 0.1s where the center if the system has moved to
 ft 
x = x0 +V0δ t = x0 + 1  (0.1s)
 s
or
x − x0 = 0.1ft
At this position, the average fluid density
slug
 1.800 slug   0.25slug 
−
 (0.1ft)=1.775 3 .
3
4
ft
ft

  ft

ρ =
Then*
d ρ ρ − ρ0
≈
=
dt
δt
(1.775 − 1.800)
slug
ft3
0.1s
dρ
slug
≈ −0.25 3
dt
ft s
*Note that this step is valid only since
constant.
∂ρ
= 0 at x0 and the mass of a system remains
∂t
Problem 4.65
3
The figure below shows a fixed control volume. It has a volume V0 = 1.0ft , a flow area
A = 1.0 ft 2 , and a length  0 = 1.0 ft . Position x represents the center of the control volume
where the fluid velocity V0 = 1.0ft/s and the density ρ0 = 1.800slug/ft . Also, at position x ,
the fluid density does not change locally with time but decreases in the axial direction at the
DV
for the system.
linear rate of 0.25 slug/ft 4. Find
Dt
3
System and control
volume at time t
ℓ0 = 1 ft
V0, ρ0
+
x0
A = 1 ft2
System at time t + st
Solution 4.65
3
GIVEN: Fixed control volume in the figure of volume V0 = 1ft . Position x0 (center of
control volume) has velocity V0 = 1ft/s, density ρ0 = 1.800slug/ft , time rate of change of
∂ρ
density
= 0, and linear spatial rate of change of density = −0.25 slug/ft 4.
∂t
3
FIND:
DV
Dt
SOLUTION: Since the system mass is constant, denoting the volume at time t + δ t by
V0 + δ V and density at time t + δ t by ρ0 + δρ , we get
ρ0V0 = ( ρ0 + δρ )(V0 + δ V ) = ρ0V0 + δρV0 + ρ0δ V + δρδ V
0 = V0δρ + ρ0δ V + δρδ V
We now divide by δ x and neglect higher order terms while letting δ x →0 to get
ρ0
dV
dρ
= −V0
dx
dx
or
−1ft3  −0.25slug 
dV −V0 d ρ
=
=

dx
ρ0 dx  1.8slug  
ft 4



3
 ft

dV
ft3
= 0.139 .
dx
ft
Then
V − V0 =
dV
( x − x0 ).
dx
Now consider a small time period δ t where ( x − x0 ) = V0δ t. Then*

dV V − V0 dV  V0δ t  dV
ft3   ft 
=
=
=
=
V
0.139
 1.0 
0 

dt
dx  δ t  dx
ft
s
δt


dV
ft3
= 0.139
dt
s
*Note that this set is valid only since
system remains constant.
∂V
∂ρ
= 0 at x0 since
= 0 at x0 and the mass of the
∂t
∂t
Problem 4.66
Water enters a 5-ft-wide, 1-ft-deep channel as shown in the figure below. Across the inlet,
the water velocity is 6 ft/s in the center portion of the channel and 1ft/s in the remainder of
it. Farther downstream, the water flows at a uniform 2 ft/s velocity across the entire
channel. The fixed control volume ABCD coincides with the system at time t = 0 . Make a
sketch to indicate (a) the system at time t = 0.5s and (b) the fluid that has entered and exited
the control volume in that time period.
A
1 ft/s
6 ft/s
1 ft
1 ft/s
D
B
2 ft
2 ft/s
5 ft
2 ft
C
Control surface
Solution 4.66
During the t = 0.5s time interval, the fluid that was along line BC at time t = 0 has moved to
the right a distance  = Vt = 2 ft/s(0.5s) = 1ft. Similarly, portions of the fluid along line AD
have moved  = 1ft/s(0.5s) = 0.5ft and  = 6 ft/s(0.5s) = 3ft. This assume the 1ft/s and 6 ft/s
fluid streams do not mix or intermingle during the 0.5 s time interval. See the figure below.
A
A'
B
B'
1 ft
Fluid that
entered
control
volume
3 ft
Fluid that
exited control
volume
0.5 ft
D
D'
C
C'
Fixed control volume
System at t = 0.5 s
Problem 4.67
The figure below illustrates a system and fixed control volume at time t and the system at a
short time δ t later. The system temperature is T = 100°F at time t and T = 103°F at time
t + δ t, where δ t = 0.1s. The system mass, m is 2.0 slugs, and 10 percent of it moves out of the
Btu
⋅ °F. The
control volume in δ t = 0.1s. The energy per unit mass u is cvT , where cv = 32.0
slug
energy U of the system at any time t is mu. Use the system or Lagrangian approach to
DU
DU
of the system. Compare this result with the
evaluated with that of the
evaluate
Dt
Dt
material time derivative and flux terms.
System
at time t
I has 10% of system mass
at time t.
I
Control
volume
II has 90% of system mass
at times t and t + δ t.
II
III
III has 10% of system mass
at time t + δ t.
System
at time t + δ t
Solution 4.67
GIVEN: System temperature is T = 100°F at time t and T = 103°F at time t + δ t, with
δ t = 0.1s. System mass is 2.0slugs. System energy per unit mass u = cvT with
Btu
cv = 32.0
⋅ °F.U = mu .
slug
dU
FIND:
using Lagrangian approach and then Eulerian time derivative plus local
dt
convective terms.
SOLUTION: The Lagrangian approach gives

Btu 
(2.0 slugs)  32.0
(103°F − 100 °F)
slug ⋅ °F 
dU U t +δ t − U t mcv (Tt +Δt − Tt )

=
=
=
dt
δt
δt
0.1s
Btu
dU
= 1920
dt
s
The Eulerian approach with b = u gives
( ρ cvTdV )II,t +Δt − ( ρ cvTdV )II,t ( mcvT )II,t +Δt − ( mcvT )II,t
d
c
TdV
ρ
=
=
v
dt V
Δt
Δt
cv


Btu 
Btu 
(0.9 × 2.0 slugs)  32.0
(103°F) − (0.9 × 2.0 slugs)  32.0
(100 °F)

slug ⋅ °F 
slug ⋅ °F 
Btu


=
= 1728
0.1s
s

Btu 
(0.1× 2.0 slugs)  32.0
(103°F)
slug ⋅ °F 
Btu



= 6592
U out =
0.1 s
s

Btu 
(0.1× 2.0 slugs)  32.0
(100 °F)
slug ⋅ °F 
Btu



U in =
= 6400
0.1s
s
and
d
Btu
ρcvTdV + U out − U in = (1728 + 6592 − 6400)

dt V
s
cv
d
Btu
ρ cvTdV + U out −U in = 1920

dt V
s
cv
Problem 4.68
The wind blows across a field with an approximate velocity profile as shown in the figure
below. Use Equation B out =  ρ bV ⋅ ndA with the parameter b equal to the velocity to
csout
determine the momentum flowrate across the vertical surface A − B , which is of unit depth
into the paper.
15 ft/s
B
20 ft
10 ft
A
Solution 4.68
B AB =  ρ bV ⋅ ndA =  ρ VV ⋅ ndA = ρ
AB
But, V =
AB
y =20 ft

y =0
20
(Vi )[(Vi ) ⋅i ](1ft)dy = ρi  V 2 dy
0
15 ft
ft
y for 0 ≤ y ≤ 10 ft (i.e., V = 0 at y = 0; V = 15 at y = 10)
10 s
s
and V = 15
ft
for y ≥10 ft
s
Thus,
2
20


3 10
10

10 
B = ρi   15 y  dy + (15)2 dy  = ρi 2.25 y
y
+
225
AB
 

0 

3
 0  10 

10
0


slugs 
ft 4
ft 4 
= 0.00238 3 750 2 + 2250 2  i
ft 
s
s 
slug ft
= 7.14i
= 7.14i lb
2
s
Problem 4.69
Water flows from a nozzle with a speed of V = 10 m/s and is collected in a container that
moves toward the nozzle with a speed of Vcv = 2 m/s as shown in the figure below. The
moving control surface consists of the inner surface of the container. The system consists of
the water in the container at time t = 0 and the water between the nozzle and the tank in the
constant diameter stream at t = 0. At time t = 0.1s , what volume of the system remains
outside of the control volume? How much water has entered the control volume during this
time period? Repeat the problem for t = 0.3s.
Stream diameter = 0.1 m
Nozzle
Container
at t = 0
Stream
at t = 0
V = 10 m/s
Vcv = 2 m/s
Moving
control
volume
3m
Solution 4.69
During the time interval δ t, the control volume moves to the left a distance of Vcvδ t = 2δ t m
and the water that was at the nozzle exit moves a distance V δ t =10δ t m to the right, as
shown below.
10 δ t
0.1 m
(3 – 12 δ t) m
2 δt
The amount of the system outside the control volume
π
= (0.1)2 (3 − 12δ t ) m3 = 0.0141m3 at t =0.1s
4
At t = 0.3s, (3 − 12δ t ) = −0.6 m
i.e., the entire system is within the control volume; none remains outside.
During the δ t = 0.1 s, time period
π
4
(0.1)2 (12δ t ) m3 = 0.00942 m3 has entered the control
volume.
During the δ t = 0.3 s, time period all of the system that was outside the control volume has
entered the control volume, i.e.,
π
4
(0.1)2 (3) m3 = 0.0236 m3
Problem 5.1
Use the Reynolds transport theorem with B = volume and, therefore,
b = volume/mass = 1/density to obtain the continuity equation for steady or unsteady
incompressible flow through a fixed control volume:  V ⋅ n dA = 0.
CV
Solution 5.1
The Reynolds Transport Theorem is
∂
ρ bd V +  ρ bV ⋅ n
dA

CV
CS
Dt
∂t
1
and with B = V and b = , we have
DBsys
=
ρ
DV sys ∂
1
1
=  ρ  d V +  ρ   V ⋅ n
dA
cs
Dt
∂t
ρ
ρ
or
DV sys ∂
= V cν +  V ⋅ n
dA
cs
Dt
∂t
(1)
For incompressible flow, the volume of the system, V sys , is constant with time so
DV sys
=0
Dt
(2)
Also, for a fixed control volume, V cv does not change and so
∂ V cv
=0
∂t
(3)
and thus combining Eqs. (1), (2), and (3), we get
cs V ⋅ n dA = 0
Problem 5.2
An incompressible fluid flows horizontally in the x–y plane with a velocity given by
y

−4 
h
 m/s, v = 0 where y and h are in meters and h is a constant. Determine
u = 30 1 − e




the average velocity for the portion of the flow between y = 0 and y = h .
Solution 5.2
y
u = 30
From the equation
 ρ V ⋅ ndA or with ρ = constant,
V= A
dy
ρA
h
 V ⋅ ndA
V= A
u=0
A
Consider a unit depth normal to the x–y plane so that
A = 1× h = h and dA = 1× dy = dy
Thus,
V=
 udA
A
or
V = 20
m
s
=
 udy
0
h
y
h
h
=
0 30 h dy
h
1
3
y =h
30 2  2 
2
m
2
= 3 y dy = 3 y   ⏐ = (30)
0
s
 3  y =0 3
h2
h2
30

h
u
u = 30 m/s
y
h
Problem 5.3
Water flows steadily through the horizontal piping system as shown in the figure below.
The velocity is uniform at section (1), the mass flowrate is 10 slugs/s at section (2), and the
D
velocity is nonuniform at section (3). (a) Determine the value of the quantity
ρd V
Dt sys
where the system is the water contained in the pipe bounded by sections (1), (2), and (3). (b)
Determine the mean velocity at section (2). (c) Determine, if possible, the value of the
integral  ρV ⋅ ndA over section (3). If it is not possible, explain what additional
(3)
information is needed to do so.
10 slugs/s
Area = 0.3 ft2
(2)
15 ft/s
(1) Area = 0.7 ft2
(3) Area = 0.7 ft2
Solution 5.3
Use the control volume that are shown with the dashed lines in the figure below.
10 slugs/s
Area = 0.3 ft2
(2)
15 ft/s
(1) Area = 0.7 ft2
(3) Area = 0.7 ft2
(a) From the conservation of mass principle, we get
D
ρ d V = 0 since  ρ d V is the unchanging mass of the system.
sys
Dt sys

(b) m2 = ρ A2V2 thus
slugs
m
ft
s
V2 = 2 =
= 17.2
slugs 
s
ρ A2 
2
1.94 3  0.3ft
ft 

10
(
)

(c) m3 =  ρ V ⋅ ndA and from the conservation of mass principle, we get
A3



m1 = m2 + m3
Thus




slugs 
slugs

m3 = m1 − m2 = ρ A1V 1 − m2 = 1.94 3  (0.7 ft 2 )(15 f/s) − 10
s
ft 


m 3 = 10.4
slugs
=  ρ V ⋅ n dA
A3
s
Problem 5.4
Water flows out through a set of thin, closely spaced blades as shown in the figure below
with a speed of V = 10 ft/s around the entire circumference of the outlet. Determine the
mass flowrate through the inlet pipe.
Inlet
0.08-ft diameter
0.1 ft
Blades
0.6 ft
60°
V = 10 ft/s
Solution 5.4
Use the control volume container within the broken lines as shown in the sketch below.
Inlet
0.08-ft diameter
0.1 ft
Blades
0.6 ft
60°
V = 10 ft/s


From the conservation of mass principle m inlet = m outlet

Also moutlet = ρ AoutletVoutlet cos 60 = ρ 2π routlet hVoutlet cos 60
slugs 

 ft 
= 1.94 3  2π ( 0.6 ft )( 0.1ft ) 10  cos 60
ft 
 s

slugs
= 3.66
s
Problem 5.6
The pump shown in the figure below produces a steady flow of 10 gal/s through the nozzle.
Determine the nozzle exit diameter, D2 , if the exit velocity is to be V2 = 100 ft/s .
Section (1)
Pump
V2
Section (2)
D2
Solution 5.6
gal 
in.3   1ft3 
ft3
For steady flow Q1 = Q2, where Q1 = 10
 231

 = 1.337
s 
gal   1728 in.3 
s
Thus, with V2 = 100
ft
,
s
ft3
π
ft 

= A2V2 = D22 100 
1.337
s
4
s

or
D2 = 0.130 ft = 1.57 in.
Problem 5.7
The fluid axial velocities shown in the figure below are the average velocities measured in
ft/s in each annular area of a duct. Find the volume flowrate for the flowing fluid.
3.2
4.4
4.7
5.0
1ʺ
+
2ʺ
2½ʺ
3ʺ
Solution 5.7
The volume flowrate is
Q =  udA


ft 
2
 5.0 s  (1in.)




 
ft  2 2
2 
4.7
in.
+
2
−1




3
4
s
π


 = 0.825 ft
Q =  udA ≈  ui Ai =
A
s

ft 
in.2   
2
2
2
j =1
 144 2   +  4.4  2.5 − 2 in. 
s
ft   


 

ft  2
2
2
 +  3.2  3 − 2.5 in. 
s
 

(
)
(
)
(
)
Problem 5.8
The human circulatory system consists of a complex branching pipe network ranging in
diameter from the aorta (largest) to the capillaries (smallest). The average radii and the
number of these vessels are shown in the table. Does the average blood velocity increase,
decrease, or remain constant as it travels from the aorta to the capillaries?
Vessel
Average Radius, mm
Number
12.5
2.0
0.03
0.006
1
159
1.4 × 107
3.9 × 109
Aorta
Arteries
Arterioles
Capillaries
Solution 5.8
The product of the square of the average radius, which is a measure of the flow area, and
the number of blood vessels is in the last column of the table.
Vessel
Aorta
Arteries
Arterioles
Capillaries
Average Radius, mm
Number
R2 Number
12.5
2.0
0.03
0.006
1
159
1.4 × 107
3.9 × 109
156
636
1.26 × 104
1.40 × 105

The average blood velocity, V , is related to the blood mass flow, m, by

m
V=
ρ AN
where ρ is blood density, A is vessel cross section area (π r 2 ) , and N is number of vessels.

So for constant m and ρ , V =

m
( ρπ r N )
2
and since the r 2 N product becomes larger, the
average velocity becomes smaller as the blood travels from the aorta to the capillaries.
Problem 5.9
Air flows steadily between two cross sections in a long, straight section of 0.1-m-insidediameter pipe. The static temperature and pressure at each section are indicated in the
figure below. If the average air velocity at section (1) is 205 m/s , determine the average air
velocity at section (2).
D = 0.1 m
Section (1)
Section (2)
p1 = 77 kPa (abs)
T1 = 268 K
V1 = 205 m/s
p2 = 45 kPa (abs)
T2 = 240 K
Solution 5.9
For steady flow between sections (1) and (2)


m1 = m2
or
ρ2A2V2 = ρ1AV
1 1
Thus
V2 =
ρ1 A1
V
ρ2 A2 1
(1)
Assuming that under the conditions of this problem, air behaves as an ideal gas, we use the
ideal gas equation of state ( p = ρ RT ) to get
ρ1 p1 T2
=
ρ2 p2 T1
(2)
Combining Eqs. (1) and (2) and observing that A1 = A2, we get
V2 =
77 kPa ( abs )  ( 240 K ) 
p1 T2
m
V1 = 
 205 
s 
p2 T1
 45 kPa ( abs )  ( 268K ) 
V2 = 314
m
s
Problem 5.10
A hydraulic jump is in place downstream from a spillway as indicated in the figure below.
Upstream of the jump, the depth of the stream is 0.6 ft and the average stream velocity is
18 ft/s . Just downstream of the jump, the average stream velocity is 3.4 ft/s . Calculate the
depth of the stream, h, just downstream of the jump.
0.6 ft
h
3.4 ft/s
18 ft/s
Solution 5.10
For steady in compressible flow between sections (1) and (2)
Q1 = Q2
or
V1A1 = V2A2
Thus
V1h1 = V2h2
and
 ft 
18
( 0.6 ft )
V1h1  s 
=
= 3.18ft
h2 =
ft 

V2
 3.4 
s

Problem 5.11
A woman is emptying her aquarium at a steady rate with a small pump. The water pumped
to a 12-in.-diameter cylindrical bucket, and its depth is increasing at the rate of 4.0 in. per
minute. Find the rate at which the aquarium water level is dropping if the aquarium
measures 24 in. (wide) × 36 in. (long) × 18 in. (high).
Solution 5.11
GIVEN: Water pumped out of aquarium at steady rate and fills 12-in.-diameter, cylindrical
bucket whose water depth increases at 4.0 in. per minute. Aquarium measures 24 in. wide
by 36 in. long by 18 in. high.
FIND: Rate at which aquarium water level is dropping.
SOLUTION: Using the attached sketch and assuming constant water density, conservation
of mass gives rate of water drained out of aquarium = rate of water filling bucket.
ha
Aa
hb
Ab
In mathematical terms
− Aa
dha
dh
= Ab b
dt
dt
where the negative sign is used since Aa , Ab , and
dhb
dh
are positive but a is negative.
dt
dt
π (12 in.)
dh
A dh
π d 2 dh
 4.0 in. 
Then a = − b b = − b b = −


dt
Aa dt
4Aa dt
4 ( 36 in. × 24 in.)  min 
2
dha
in.
= −0.524
dt
min
Problem 5.12
An evaporative cooling tower (see the figure below) is used to cool water from 110 to 80 F.
Water enters the tower at a rate of 250,000 lbm/hr . Dry air (no water vapor) flows into the
tower at a rate of 151,000 lbm/hr . If the rate of wet airflow out of the tower is
156,900 lbm/hr , determine the rate of water evaporation in lbm/hr and the rate of cooled
water flow in lbm/hr.
Wet
air out
Water in
Air in
Air in
Water out
Solution 5.12
Wet air
m∙ = 156,900 Ibm/hr
For steady flow of dry air


m3 = m2,dry air
(1)
Warm water
For steady flow of water


m∙ = 250,000 Ibm/hr
(1)
Dry air
(3)
(2)

m1 = m 2,water + m 4
(2)
Also



m2 = m2,dry air + m2,water
(3)
Combining Eqs. (1) and (3), we get



m 2,water = m 2 − m3 = rate of water evaporation
So

m2,water = 156900
lbm
lbm
lbm
− 151000
= 5900
hr
hr
hr
From Eq. (2), we get



m 4 = m1 − m 2,water = rate of cooled water flow
or

m4 = 250000
lbm
lbm
lbm
− 5900
= 244000
hr
hr
hr
m∙ = 151,000 Ibm/hr
(4)
Cooled
water
Problem 5.13
At cruise conditions, air flows into a jet engine at a steady rate of 65 lbm/s . Fuel enters the
engine at a steady rate of 0.60 lbm/s . The average velocity of the exhaust gases is 1500 ft/s
relative to the engine. If the engine exhaust effective cross-sectional area is 3.5 ft 2 , estimate
the density of the exhaust gases in lbm/ft 3 .
Solution 5.13
For steady flow


Fuel in

m3 = m1 + m2
Section (2)
or
Section (3)


ρ3 A3V3 = m1 + m 2
Air in
Thus
Exhaust
gases out


m +m
ρ3 = 1 2 =
A3V3
ρ3 = 0.0125
lbm
lbm
+ 0.60
s
s
ft


(3.5 ft 2 ) 1500 
s


65
lbm
ft3
Section (1)
Control
volume
Problem 5.14
Water at 0.1m 3 /s and alcohol (SG = 0.8) at 0.3 m 3 /s are mixed in a y-duct as shown in the
figure below. What is the average density of the mixture of alcohol and water?
Water and
alcohol mixture
Water
Q = 0.1 m3/s
Alcohol (SG = 0.8)
Q = 0.3 m3/s
Solution 5.14
Water and
alcohol mixture
(3)
(1)
Water
Q = 0.1 m3/s
(2)
Alcohol (SG = 0.8)
Q = 0.3 m3/s
For steady flow



m1 + m2 = m3
or
ρ1Q1 + ρ 2Q2 = ρ3Q3 (1)
Also, since the water and alcohol may be considered incompressible
Q1 + Q 2 = Q3 (2)
Combining Eqs. (1) and (2), we get
ρ1Q1 + ρ2Q2 = ρ3 (Q1 + Q2 )
or
ρ3 =
ρ1Q1 + ρ2Q2
Q1 + Q2
and
ρ3 = ρ1
(Q1 + SG Q2 )
Q1 + Q2
Thus
ρ3 =

kg   m3
m3  

+ ( 0.8 )  0.3

 999 3  0.1

s
s  
m  


3
0.1
m
m
+ 0.3
s
s
3
= 849
kg
m3
Problem 5.15
In the vortex tube shown in the figure below, air enters at 202 kPa absolute and 300 K .
Hot air leaves at 150 kPa absolute and 350 K , whereas cold air leaves at 101kPa absolute


and 250 K . The hot air mass flowrate, mH , equals the cold air mass flowrate, m C . Find
the ratio of the hot air exit area to cold air exit area for equal exit velocities.
TC = 250 K
pC = 101 kPa
m∙ C
TH = 350 K
pH = 150 kPa
m∙ H
T = 300 K
p = 202 kPa
Solution 5.15


GIVEN: Vortex tube in figure. mC = mH .
Equal exit velocities, VC = VH .
FIND: Ratio of hot exit area to cold exit area.


mC
mH
SOLUTION: The two exit areas are AH =
and AC =
ρHVH
ρCVC


Since mC = mH and VC = VH ,
AH ρC
=
.
AC ρH
Assuming the air is an ideal gas,
AH  PC   RTH   PC   TH   101kPa   350 K 
=

=
=



AC  RTC   PH   PH   TC   150 kPa   250 K 
AH
= 0.943
AC
Problem 5.16
Molten plastic at a temperature of 510 °F is augured through an extruder barrel by a screw
occupying 3/5 of the barrel’s volume (see figure below). The extruder is 16 ft long and
has an inner diameter of 8 in . The barrel is connected to an adapter having a volume of
0.48 ft3 . The adapter is then connected to a die of equal volume. The plastic exiting the die
is immediately rolled into sheets. The line is producing 4-ft widths of material at a rate of
30 ft/min and a gauge thickness of 187 mil . What is the axial velocity, V1, of the plastic in
the barrel? Assume that the plastic density is constant as it solidifies from a liquid (in the
extruder) into a solid sheet.
Rolls
Die Adapter
Extruder
barrel
V2 =
30 ft/min
Screw
V1
t
Thin sheet
Solution 5.16
Apply conservation of mass (i.e., continuity equation) to a control volume enclosing the
plastic. Assume steady state.
A2
V1
V2
A1
V1
For constant density (and steady state)
AV
1 1 = A2V2
Now
2
 ft 
2π
2π
2
A1 =
D=
(8in.)2 
 = 0.1393ft ,
54
54
 12 in. 
 0.187 
ft  = 0.0623ft 2 ,
A2 = bt = (4 ft) 
 12

and
ft 

0.0623ft 2 )  30
(

AV
 min  .
V1 = 2 2 =
A1
( 0.1396 ft2 )
V1 = 13.4
ft
min
Problem 5.17
A water jet pump (see the figure below) involves a jet cross-sectional area of 0.01m 2 , and a
jet velocity of 30 m/s . The jet is surrounded by entrained water. The total cross-sectional
area associated with the jet and entrained streams is 0.075 m 2 . These two fluid streams
leave the pump thoroughly mixed with an average velocity of 6 m/s through a crosssectional area of 0.075 m 2 . Determine the pumping rate (i.e., the entrained fluid flowrate)
involved in l/s .
6 m/s
Entrained
water
30 m/s
jet
Entrained
water
Solution 5.17
6 m/s
(2)
Entrained
water
30 m/s
jet
(1)
(3)
Entrained
water
For steady incompressible flow through the control volume
Q1 + Q2 = Q3
or
V1A1 + Q2 = V3A3
Thus
 m 

l 
 m
Q2 = V3 A3 −V1A1 =  6  0.075 m 2 −  30  0.01m 2  1000 3 
s 
m 

 s 

l
Q2 = 150
s
(
)
(
)
Problem 5.18
To measure the mass flowrate of air through a 6-in.- inside diameter pipe, local velocity
data are collected at different radii from the pipe axis (see the table below). Determine the
mass flowrate corresponding to the data listed in the following table. Plot the velocity
profile and comment.
r (in.)
Axial Velocity (ft/s)
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
1.6
1.8
2.0
2.2
2.4
2.6
2.8
2.9
2.95
2.98
3.00
30
29.71
29.39
29.06
28.70
28.31
27.89
27.42
26.90
26.32
25.64
24.84
23.84
22.50
20.38
18.45
16.71
14.66
0
Solution 5.18
The mass flowrate is calculated with

R
R
0
0
m =  ρ u 2π rdr = 2πρ  urdr
where
R = 3 in.
ρ = 0.00238
slug
ft 3
u = local axial velocity in
ft
s
r = local radius in in.
R
and  urdr is evaluated numerically with the trapezoidal rule with unequal intervals.
0
The result is:
r=
3
ft
12

slug 
ft
slugs

m = 2π  0.00238 3   u ( r ft )( dr ft ) = 0.0114
s
ft  r =0 s

Problem 5.19
Two rivers merge to form a larger river as shown in the figure below. At a location
downstream from the junction (before the two streams completely merge), the nonuniform
velocity profile is as shown and the depth is 6 ft . Determine the value of V .
0.8 V
30 ft
70 ft
3 ft/s
V
50 ft
Depth = 3 ft
80 ft
Depth = 5 ft
4 ft/s
Solution 5.19
0.8 V
30 ft
70 ft
3 ft/s
(3)
50 ft
V
Depth = 3 ft
(1)
(2)
80 ft
Depth = 5 ft
4 ft/s
Use the control volume shown within broken lines in the sketch above. We note that

m = ρ AV and from the conservation of mass principle, we get
m 1+ m 2 = m 3 = m 3,0.8V + m 3,V
Thus
ρ AV
1 1 + ρ A2V2 = ρ A0.8V 0.8V + ρ AV V
and
AV
1 1 + A2V2
=
V=
A0.8V ( 0.8 ) + AV
V = 3.63
ft
s
 ft 
 ft 
(50 ft)( 3 ft)  3  + (80 ft)( 5 ft)  4 
 s
 s
(30 ft)( 6 ft)(0.8) + (70 ft)( 6 ft)
Problem 5.20
Various types of attachments can be used with a shop vac. Two such attachments are
shown in the figure below—a nozzle and a brush. The flowrate is 1ft3 /s . (a) Determine the
average velocity through the nozzle entrance, V n . (b) Assume the air enters the brush
attachment in a radial direction all around the brush with a velocity profile that varies
linearly from 0 to V b along the length of the bristles as shown in the figure. Determine the
value of V b .
Q = 1 ft3/s
Q = 1 ft3/s
1.5 in.
2-in. dia.
Vb
Vn
3-in. dia.
Solution 5.20
(a) Q1 = Q2 where Q2 = 1
ft3
s
Thus,
ft3
1
s
=
≡
=
AV
Q
or
V
V
1 1
2
1
n
2
π2 
ft


4  12 
so
Vn = 45.8
ft
s
(b) Q3 = Q4 where Q4 = 1
ft3
and Q3 = V3A3 where
s
1
V3 = average velocity at (3) = Vb and A3 = π D3 h3
2
Thus,
1   3   1.5  
ft 3
Vb π  ft  
ft   = 1 , or
2   12   12  
s
Vb = 20.4
ft
s
Problem 5.21
An appropriate turbulent pipe flow velocity profile is
1n
R−r  
V = uc 
 i
 R 
where uc = centerline velocity, r = local radius, R = pipe radius, and i = unit vector along
pipe centerline. Determine the ratio of average velocity, u, to centerline velocity, uc , for (a)
n = 4, (b) n = 6, (c) n = 8, (d) n = 10. Compare the different velocity profiles.
Solution 5.21
For any cross section area
Cross section
area

m = ρ Au =  ρ V ⋅ ndA
dr
Also
r
1
R
 R − r n
V ⋅ n = V ⋅i = uc 

 R 
Thus for a uniformly distributed density, ρ , over area A
dA = 2π rdr
1
 R − r n
R
 uc  R  2π rdr
u= 0
π R2
and
1
r n  r 
R
u
2 n2

r
= 2  1 −    d   = 2
uc
R   R   R  2 n + 3n + 1
0
1
n
u
uc
0.9
4
0.711
0.8
0.7
0.791
u/uc
0.6
6
0.5
0.4
8
0.837
0.3
0.2
0.1
10
0.866
0
0
2
4
6
8
10
The different velocity profiles (including for laminar flow) are compared in the figure
below.
1.0
n = 10
n=6
Laminar
n=8
r
0.5
R
Turbulent
0
0
0.5
1.0
u
Vc
Since the profile for n = 4 is not found in wholly turbulent flows, it is not shown.
Problem 5.22
As shown in the figure below, at the entrance to a 3-ft-wide channel, the velocity
distribution is uniform with a velocity V . Further downstream, the velocity profile is given
by u = 4 y − 2 y2, where u is in ft/s and y is in ft. Determine the value of V .
V
0.75 ft
y
u = 4y – 2y2
1 ft
x
Solution 5.22
(2)
(1)
V
0.75 ft
y
1 ft
u = 4y – 2y2
x
Use the control volume indicated by the broken lines in the sketch above.
From the conservation of mass principle
Q1 = Q2
1ft
V1A1 =  udA =  ( 4 y − 2 y2 ) b dy
A2
0
1ft
 4 y2 2 y3 
4b ft3
V ( 0.75 ft ) b = 3 
−
 b=
3 
3 s
 2
0
V=
4
ft
= 1.78
3(0.75)
s
Problem 5.23
The cross-sectional area of a rectangular duct is divided into 16 equal rectangular areas, as
shown in the figure below. The axial fluid velocity measured in feet per second in each
smaller area is given in the figure. Estimate the volume flowrate and average axial velocity.
20.0 in.
3.0
3.4
3.6
3.1
3.7
4.0
3.9
3.8
3.9
4.6
4.5
4.2
3.7
4.4
4.3
3.9
16.0 in.
Velocities in ft/s
Solution 5.23
GIVEN: Axial velocities measured in rectangular duct shown in the figure in the problem.
FIND: Estimate volume flowrate and average velocity.
SOLUTION:
Q =  V ⋅ ndA
A
The velocity in each small rectangular area is assumed to be the average velocity for that
area.
16
16
i =1
i =1
Then Q = Vi Ai = Ai Vi
since Ai is the same value for all areas.
Adding the velocities in the figure gives
16
ft
Vi = 62.0 s
i =1
so
ft   ft 2 

Q = ( 5 in. × 4 in.)  62.0  

s   144 in.2 

or
ft3
Q = 8.61
s
The average velocity for the entire duet is

ft3 
 8.61 
s 
Q

V= =
or
A
 ft 2 
(20.0 in. ×16.0 in.) 
 144 in.2 


V = 3.87
ft
s
Problem 5.24
Oil for lubricating the thrust bearing shown in the figure below flows into the space
between the bearing surfaces through a circular inlet pipe with velocity
  r 2 
u = U 0 1 −   ,
  R  
where R = 1.5 mm . The oil has a specific gravity SG = 0.86 and flows in the inlet pipe
at a rate of 0.006 kg/s. Compute the average velocity V1 of the oil in the inlet pipe and
the average velocity V 2 at the outlet (plane 2) and the maximum velocity in the oil inlet
pipe (U 0 ). Assume radial flow.
D2 = 10 cm
Circular disc
Outlet
plane 2
D1 = 3.0 mm = 2R
Inlet
pipe
u = U0 1 –
r
h = 2.0 mm
(Rr (
2
= oil velocity profile
Solution 5.24
  r 2 
GIVEN: Inlet oil velocity = u = U 0 1 −   , oil flowrate 0.006 kg/s, oil specific gravity =
  R  
S = 0.86 , and R = 1.5 mm .
FIND: Average velocity V1 in inlet pipe, average velocity V 2 in outlet plane, and maximum
velocity U 0 .
SOLUTION: To find the average velocity V1, we use the flowrate Q
2
kg 
mm 

4  0.006 1000

Q
m
4m
s 
m 

=
=
=
V1 =
kg
A1 ρ A1 SG ρ waterπ D12
( 0.86 ) 1000 3  π ( 3.0 mm )2

m 

V1 = 0.987

m
s
The velocity V 2 is found from
2
kg 
cm 
mm 

0.006 100
1000



Q
m
m
s 
m 
m 

=
=
=
V2 =
kg
A2 ρ A2 SG ρwaterπ D2 h
( 0.86 ) 1000 3  π (10 cm)(2.0 mm)2

m 
m
V1 = 0.0111
s


The velocity U 0 is found by first finding the volume flowrate
R
R
  r 2 
Q =  2π ru dr =  2π rU 0 1 −    dr
  R  
0
0
R

 r2
r3 
r4 
= 2πU 0   r − 2  dr = 2πU 0  −
2
 R 
2
4
R

 0



0
R
 R 2 R 4  π R 2U 0
= 2πU 0 
−
=
4 
2
 2
So
2
kg  
mm 

2  0.006  1000

2Q
2m
2m
s 
m 

U0 =
=
=
=
π R 2 ρπ R 2 SG ρ waterπ R 2 (0.86) 1000 kg  π (1.5 m)2


m3 


U 0 = 1.97
m
s

Problem 5.25
Flow of a viscous fluid over a flat plate surface results in the development of a region of
reduced velocity adjacent to the wetted surface as depicted in the figure below. This region
of reduced flow is called a boundary layer. At the leading edge of the plate, the velocity
profile may be considered uniformly distributed with a value U . All along the outer edge of
the boundary layer, the fluid velocity component parallel to the plate surface is also U . If
the x -direction velocity profile at section (2) is
3
u 3 y 1 y
=  −  
U 2δ  2δ 
develop an expression for the volume flowrate through the edge of the boundary layer from
the leading edge to a location downstream at x where the boundary layer thickness is δ .
Section (2)
Outer edge
of
boundary
layer
U
Section (1)
U
δ
x
Solution 5.25
Consider the control volume shown. Apply
the Conservation of Mass principle to get
Section (2)
U
m bl = m 2 =  ρ V nˆ dA
Outer edge
of
boundary
layer
U
Section (1)
Control
volume
m∙ bl
A2
m∙ 2
For incompressible flow,
1
3
3 y 1 y   y
ρQbl = ρU δ     −   d  
0
 2  δ  2  δ    δ 
where  = width of the plate. Therefore
1
 3  y 2 1  y  4 
5
Qbl = ρU δ    −    = U δ
 4  δ  8  δ   0 8
x
δ
Problem 5.26
Air at standard conditions enters the compressor shown in the figure below at a rate of
3
10 ft 3 /s . It leaves the tank through a 1.2-in.-diameter pipe with a density of 0.0035slugs/ft
and a uniform speed of 700 ft/s . (a) Determine the rate (slugs/s) at which the mass of air in
the tank is increasing or decreasing. (b) Determine the average time rate of change of air
density within the tank.
Compressor
Tank volume = 20 ft3
10 ft3/s
1.2 in.
700 ft/s
0.0035 slugs/ft3
0.00238 slugs/ft3
Solution 5.26
Compressor
Tank volume = 20 ft3
10 ft3/s
1.2 in.
700 ft/s
0.0035 slugs/ft3
0.00238 slugs/ft3
Use the control volume within the broken lines.
(a) From the conservation of mass principle, we get
DM sys
Dt
DM sys
Dt
DM sys
Dt
DM sys
b)
Dt


= min − m out = ρinQin − ρ out AoutVout
slug   ft3  
slug  π (1.2 in.) 
ft3 

=  0.00238 3  10

 − 0.0035 3 
 700
s  
s 
ft  
ft  

in.2  
144 2 
ft 

= 0.00456
slug
s
(
) =V
=
D ρVsys
Dt
which is positive, showing that the mass is increasing
sys
slug
Dρ
= 0.00456
s
Dt
so
Dρ
=
Dt
slug
slug
0.00456
s =
s = 2.28 × 10 −4 slug
3
Vsys
20 ft
ft 3 ⋅ s
0.00456
Problem 5.27
Estimate the time required to fill water in a cone-shaped container (see the figure below)
5 ft high and 5 ft across at the top if the filling rate is 20 gal/min.
5 ft
5 ft
Solution 5.27
5 ft
Control
volume
5 ft
From application of the conservation of mass principle to the control volume shown in the
figure, we have
∂
dA = 0
 ρ d V +  ρ V ⋅ n
∂t CV
CS
For incompressible flow
∂V
−Q = 0
∂t
or
V
t
0
0
 d V = Q  dt
Thus
2
t=
V π D2 h
=
=
Q 12 Q
and
t = 12.2 min

π ( 5 ft ) ( 5 ft ) 1728
in.3 

ft3 

gal  
in.3 
231
(12 )  20



gal 
 min  
Problem 5.29
A water jet leaves a fixed nozzle with a velocity of 10 m/s. The jet diameter is 10 cm. A 30°
cone is pushed into the water jet at a speed of 5 m/s. The water impinges on the cone with
the jet axis and the cone axis in perfect alignment so that the water is divided evenly by the
cone. Bernoulli’s equation suggests that because the pressure on the jet boundary is
constant, the water velocity relative to the cone surface is constant. Determine the thickness
of the water stream when it reaches the base of the cone.
10 cm
30°
5 m/s
10 m/s
20 cm
Solution 5.29
Consider a control volume attached to the cone. The water jet approaches this control
volume at 15 m/s. The continuity equation gives V1A1 = V2 n A2 where location “1” is the
incoming jet, location “2” is the flow leaving at the base of the cone, and A2 is the flow area
in the plane of the cone base. (Note: the 20 cm dimension is assumed to be from the tip of
the cone to the base of the cone). Then
2
V1π r12 = V1 cos θπ ( r2 + δ ) − r22 


where δ is the thickness of the water stream in the radial direction. Expanding and
simplifying leads to
δ 2 + 2r2δ −
r12
=0
cos(θ )
For the situation in Fig. 5.29, r1 = 0.05 m r2 = 0.20sin(15 ) = 0.05176 m θ = 15 and the
equation becomes
δ 2 + 0.1035δ − 0.002588 = 0
Solving by the quadratic formula and discarding the negative value gives
δ = 0.0208 m = 2.08 cm
The thickness of the water stream perpendicular to the surface of the cone is
t = δ cos(θ ) = 0.0208cos(15 ) = 0.02009m = 2.009cm
Problem 5.30
A hypodermic syringe (see the figure below) is used to apply a vaccine. If the plunger is
moved forward at the steady rate of 20 mm/s and if vaccine leaks past the plunger at 0.1 of
the volume flowrate out the needle opening, calculate the average velocity of the needle exit
flow. The inside diameters of the syringe and the needle are 20 mm and 0.7 mm.
Qleak
Qout
Solution 5.30
Deforming control
volume
Qleak
Qout
Using a deforming control volume and the conservation of mass principle
DM sys
Dt
=
∂
dA = 0
 ρ d V +  ρ W ⋅ n
∂t CV
CS
−ρ AV
1 p + ρQ2 + ρQleak = 0 (1)
we obtain
Since ρ = constant, Qleak = 0.1Q2 and Q2 = A2V2
we obtain from Eq. (1)
1.1A2V2 = AV
1 p
or
 A  Vp  d12  V p
= 2 
=
V2 =  1 
 
 A2  1.1  d2  1.1
and
V2 = 14.8
m
s
( 20 mm )2  20
mm 

s 

mm 
( 0.7 mm )2 (1.1) 1000

m 

Problem 5.31
The figure below shows a two-reservoir water supply system. The water level in reservoir 1
drops at the rate of 0.01m/min , and the water level in reservoir 2 drops at the rate of
0.015m min . Calculate the average velocity V3 in the 0.50-m-diameter pipe.
A1 = 5 acres
h1
A2 = 4 acres
Reservoir 2
Reservoir 1
h2
d3 = 0.50 m
V3
Solution 5.31
GIVEN: The figure with
dh1
dh
0.01m
0.015 m
.
and 2 = −
=−
dt
min
dt
min
FIND: Average velocity in 0.50 m diameter pipe.
SOLUTION:
Assume constant water density and apply conservation of mass to a control volume
enclosing the two tanks and the 0.50 m discharge pipe.


d
( MCV ) = min − mout
dt
ρπ d3
d
ρ [ A1h1 + A2 h2 ] = 0 −
V3
4
dt
dh2  −π d32
 dh1
+
A
A
V3
2
 1
=
4
dt
dt 

dh 
−4  dh1
V3 =
A
+ A2 2 
2  1 dt
dt 
π d3 
The numerical values give
V3 =
2

 −0.01m 
 −0.015 m   1233 m
+
4
acres
5acres




2 
 min 
 min   acres
π ( 0.50 m ) 
V3 = 722
−4
m
m
= 12
min
s
Problem 5.32
The Hoover Dam backs up the Colorado River and creates Lake Mead, which is
approximately 115 miles long and has a surface area of approximately 225 square miles. If
during flood conditions, the Colorado River flows into the lake at a rate of 45,000 cfs and
the outflow from the dam is 8000 cfs , how many feet per 24-hour day will the lake level
rise?
Solution 5.32
For the control volume shown:


min − mout =
∂
ρd V
∂t CV water
Alake
Qin
h

or since m = ρQ
Qin − Qout =
Control surface
∂
∂h
( Alake h ) = Alake
∂t
∂t
Thus,
ft3
∂h Qout − Qin
s = 5.90 × 10 −6 ft
=
=
2
∂t
s
Alake
ft 
2
225 mi  5280 
mi 

( 45000 − 8000 )
= 5.90 ×10 −6
ft 
s 
hr 
ft
 = 0.510
 3600   24
s
hr   day 
day
Lake Mead
Qout
Problem 5.33
Storm sewer backup causes your basement to flood at the steady rate of 1 in. of depth per
hour. The basement floor area is 1500 ft 2 . What capacity (gal min) pump would you rent
to (a) keep the water accumulated in your basement at a constant level until the storm
sewer is blocked off, and (b) reduce the water accumulation in your basement at a rate of
3 in. hr even while the backup problem exists?
Solution 5.33
Deforming
control volume that
contains water
h
Floor area = A
Flow out
Flow in
For a deforming control volume that contains the water over the basement floor (see sketch
DM sys ∂
dA = 0 leads to
=
above), the conservation of mass principle
 ρ d V +  ρ W ⋅ n
Dt
∂t CV
CS
∂
dA = 0
 ρ d V +  ρ V ⋅ n
∂t CV
CS
or for constant fluid density and area (A)
A
dh
− Qin + Qout = 0
dt
(1)
(a) For part a, Eq. (1) leads to
Qout = Qin
To evaluate Qin, we use Eq. (1) with Qout = 0 .
Thus



dh
1 
ft3
2  in. 
Qin = A = 1500 ft 1  
 = 125
dt
hr
 hr   12 in. 
 ft 
and
(
)



ft3  
gal   1 
gal
Qout = 125
= 15.6
  7.48 3  


hr  
min
ft   60 min 

hr 

(b) For part b, Eq. (1) yields
Qout = Qin − A
dh
dt








gal
in.
1
gal
1




Qout = 15.6
7.48 3  
− 1500 ft 2  −3  



min
 hr   12 in.  
ft   60 min 

ft 
hr 

gal
Qout = 62.4
min
(
)
Problem 5.34
“Green” 1.6-gpf standards Toilets account for approximately 40% of all indoor household
water use. To conserve water, the new standard is 1.6 gallons of water per flush (gpf). Old
toilets use up to 7 gpf; those manufactured after 1980 use 3.5 gpf. Neither is considered a
low-flush toilet. A typical 3.2-person household in which each person flushes a 7-gpf toilet
4 times a day uses 32,700 gallons of water each year; with a 3.5-gpf toilet, the amount is
reduced to 16,400 gallons. Clearly the new 1.6-gpf toilets will save even more water.
However, designing a toilet that flushes properly with such a small amount of water is not
simple. Today there are two basic types involved: those that are gravity powered and those
that are pressure powered. Gravity toilets (typical of most currently in use) have rather long
cycle times. The water starts flowing under the action of gravity and the swirling vortex
motion initiates the siphon action that builds to a point of discharge. In the newer pressureassisted models, the flowrate is large, but the cycle time is short and the amount of water
used is relatively small. (See Problem 5.34.)
When a toilet is flushed, the water depth, h, in the tank as a function of time, t, is as given in
the table. The size of the rectangular tank is 19 in. by 7.5 in. (a) Determine the volume of
water used per flush, gpf . (b) Plot the flowrate for 0 ≤ t ≤ 6 s .
t (s)
h (in.)
0
0.5
1.0
2.0
3.0
4.0
5.0
6.0
5.70
5.33
4.80
3.45
2.40
1.50
0.75
0
Solution 5.34
Volume of water per flush = 5.70 in.(19 in. × 7.5 in.) = 812 in.3
 1gal 
= 812 in.3 
= 3.52 gal
 231in.3 


a)
dh
d (volume in tank)
dh
, where
is obtained by numerical differentiation
= Atank
dt
dt
dt
of the h versus t data shown below.
b) Q =
The resulting Q versus t result are also shown below.
6
5
h (in.)
4
3
2
1
0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
t (s)
0
Q (in.3/s)
–50
–100
–150
–200
–250
0
1
2
3
t (s)
4
5
6
Problem 5.35
Find the force components Fx and Fy required to hold the box as shown in the figure below
stationary. The fluid is oil and has specific gravity of 0.85 . Neglect gravity effects.
Atmospheric pressure acts around the entire box. Steady flow conditions prevail.
Q3 = 3.0 m3/s
V3 = 1.5 m/s
V2 = 2.0 m/s
θ 3 = 60°
θ2 =
45°
Q1 = 5 m3/s
V1 = 2.0 m/s
Fx
Fy
Solution 5.35
Q3, V3
Q2, V2
θ3
θ2
Q1, V1
Fx
x
Fy
x
GIVEN: Fluid is oil with SG = 0.85. Neglect gravity effect. Atmospheric pressure acts over
entire box steady state.
FIND: Force components Fx and Fy hold box stationary.
SOLUTION:
Consider the control volume enclosing the entire box. Consider the positive directions as
shown. Write the continuity equation for this control volume. For constant density
Q1 = Q2 + Q3
or Q2 = Q1 − Q3 = 5.0
m3
m3
m3
− 3.0
= 2.0
.
s
s
s
Now write the linear momentum equation in the x-direction for the control volume.


 Fx = M x,out − M x,in
where
 Fx = Fx ,

M x,out = ρQ3 (V3 cos θ3 ) + ρQ2 ( −V2 cos θ 2 )
and

M x,in = ρQ1 ( −V1 )
So the linear momentum equation is
Fx = − ρQ3V3 cos θ3 − ρQ2V2 cos θ2 + ρQV
1 1 = ρ (QV
1 1 − Q3V3 cos θ3 − Q2V2 cos θ 2 )
 N ⋅ s2

kg  
m3   m  
m3   m 
m3  
m

= 1000 × 0.85 3   5.0
−
°
−
°
20
3.0
1.5
cos60
2.0
2.0
cos
45













s  
s  
s  
s 
s  
s 
m  


 kg ⋅ m
Fx = 4180 N
Writing the linear momentum equation is the y-direction gives


 Fy = M y,out − M y,in
where
 Fy = Fe ,

M y,out = ρQ3 (V3 sin θ3 ) + ρQ2 ( −V2 sin θ2 )
and

M y,in = ρQ1 ( 0 ) = 0
so the y linear momentum equation is
Fy = ρ (Q3V3 sin θ3 + Q2V2 sin θ2 )
2

kg  
m3   m 
m3  
m

 
 N ⋅s
= 1000 × 0.85 3   3.0
 1.5  sin 60 +  2.0
  2.0  sin 45 
s  
s 
s  
s 

m  

 kg ⋅ m
Fy = 5720 N
Problem 5.37
Find the horizontal and vertical forces to hold stationary, the nozzle as shown in the figure
m
below. The fluid flowing through it is 10 °C liquid water; A1 = 1.0 m 2 , A2 = 0.25 m 2 ,V1 = 20 ,
s
p2 = patm and p1 = patm + 30 kPa . Neglect gravity.
45°
A2, p2 = patm
Fx
V1 = 20 m/s
y
A1
Fy
x
patm
p1
Solution 5.37
GIVEN: 10 °C liquid water, A1 = 1.0 m 2 , A2 = 0.25 m 2 ,V1 = 20
m
s
V2
p2 = patm and p1 = patm + 30 kPa . Neglect gravity.
θ
FIND: Fx and Fy
SOLUTION: Apply the linear momentum equation in the xdirection to the control volume shown on the ring.


M x,out − M x,in =  Fx
where


M x,out = mV2 cos θ2 = ρ AVV
1 1 2 cos θ ,


2
M x,in = mV1 = ρ AV
1 1 ,
and
 Fx = Fx + ( p1 − patm )A1
where p1 is an absolute pressure.
The x-direction linear momentum equation is
Fx = ρ AV
1 1 (V2 cos θ −V1 ) − ( p1 − patm ) A1
Fx
V1
Fy
Assuming constant fluid density, the continuity equation gives
ρ AV
1 1 = ρ A2V2 cos θ , V2 =
AV
1 1
A2 cos θ
or
 m
(1.0 m 2 )  20 
m
s 

V2 =
= 113 .
2

s
(0.25 m ) cos 45
kg
where ρ = 1000 3 and
m
kg 
m
m   N ⋅ s2  
 m 

3 N 
2
Fx = 1000 3  1.0 m 2  20 113  cos 45 − 20  
 −  30 × 10
 1.0 m
2

s 
s 
s   kg ⋅ m  

m 
m 

(
)
(
Fx = 1.17 × 106 N = 1170 kN
We now apply the linear momentum equation in the y-direction to the control volume.


M y,out − M y,in =  Fy
where


M y,out = mV2 sin θ = ρ AVV
1 1 2 sin θ ,


M y,in = m(0) = 0,
and
 Fy = Fy
The y-direction linear momentum equation is
Fy = ρ AVV
1 1 2 sinθ .
The numerical values give
2
kg 
m 
m

2 
  N ⋅s 
Fy = 1000 3  1.0 m  20 113  sin 45 
 kg ⋅ m 
s 
s 


m 


(
)
Fy = 1.60 × 106 N = 1600 kN
)
Problem 5.38
Water flows through a horizontal bend and discharges into the atmosphere as shown in the
figure below. When the pressure gage reads 10 psi, the resultant x-direction anchoring
force, FAx, in the horizontal plane required to hold the bend in place is shown on the figure.
Determine the flowrate through the bend and the y-direction anchoring force, FAy, required
to hold the bend in place. The flow is not frictionless.
10 psi
Bend
Q=?
Water
FAx = 1440 lb
Area = 0.2 ft2
FAy = ?
Area = 0.1 ft2
45°
Solution 5.38
Control
volume
(1)
10 psi
Bend
p1A1
Q=?
Water
FAx = 1440 lb
Area = 0.2 ft2
FAy = ?
Area = 0.1 ft2
(2)
45°
A control volume that contains the bend and the water within the bend between sections (1)
and (2) as shown in the sketch above is used. Application of the x-direction component of
the linear momentum equation yields
−u1ρQ −V2 cos 45° ρQ = p1A1 − FAx + p20 gage A2 cos 45°
with
u1 =
Q
A1
and V2 =
Q
A2
Equation (1) becomes
Q 2 ρ Q 2 ρ cos 45°
−
−
= p1A1 − FAx
A1
A2
(1)
or for part (a)
Q=
Q=
− p1A1 + FAx
 cos 45 1 
ρ
+ 
A1 
 A2
lb  
in.2 

− 10 2   144 2  0.2 ft 2 + 1440 lb
ft 
 in.  
slugs   lb ⋅ s 2   cos 45
1 

+

1.94 3   1 slug ⋅ ft  
2
ft  
0.2 ft 2 

  0.1ft
(
Q = 7.01
)
ft3
s
For part (b), we use the y-direction component of the linear momentum equation to get
FAy = V2 sin 45 ρQ =
Q
sin 45 ρQ
A2
or
FAy =
Q2
sin 45 ρ
A2
and

ft 2 
 7.01 
s 
slugs   lb ⋅ s2 


FAy =
sin 45 1.94 2  1
 = 674 lb
ft   slug ⋅ ft 

0.01ft 2
(
)
Problem 5.39
Find the magnitude of the force F required to hold the plate in the figure below stationary.
V2
V1 = 30 m/s
Water
m∙ = 5 kg/s
45°
F
V3
Solution 5.39
V2
GIVEN: Plate hit with water in the figure in the
problem. Neglect gravity. Frictionless plate.
FIND: Force F to hold plate stationary.
SOLUTION: Assume steady flow and apply the
linear momentum equation in the x-direction to the
control volume shown. F is the force on the plate.


θ
y
V1
M x,out −M x,in =  Fx
where

M x,out = 0
V3

M x,in = ρ AV
1 1 (V1 sin θ ),
and
 Fx = −F .
The x-direction linear momentum equation is then
2
0 − ρ AV
1 1 sin θ = − F

2
or F = ρ AV
1 1 sin θ = mV1 sin θ
The numerical values give
 N ⋅ s2 
 kg  m 
F =  5  30  sin 45 
 kg ⋅ m 
s 
 s 


F = 106 N
F
θ
θ = 45°
x
Problem 5.40
Water enters the horizontal, circular cross-sectional, sudden contraction nozzle sketched in
the figure below at section (1) with a uniformly distributed velocity of 25 ft/s and a pressure
of 75 psi . The water exits from the nozzle into the atmosphere at section (2) where the
uniformly distributed velocity is 100 ft/s . Determine the axial component of the anchoring
force required to hold the contraction in place.
Section (2)
p2 =
0 psi
D1 = 3 in.
p1 = 75 psi
V1 = 25 ft/s
V2 =
100 ft/s
Section (1)
Solution 5.40
Section (2)
D1 = 3 in.
p1A1
p2A2
p1 = 75 psi
F
V1 = 25 ft/s A
p2 =
0 psi
V2 =
100 ft/s
Section (1)
For this problem, we include in the control volume, the nozzle, as well as the water
at an instant between sections (1) and (2) as indicated in the sketch above. The horizontal
forces acting on the contents of the control volume are show in the sketch. Note that
the atmospheric forces cancel out and are not shown. Application of the horizontal or
∂
x-direction component of the linear momentum equation
dA =
 Vρ d V +  Vρ V ⋅ n
∂t CV
CS
 Fcontents of the to the flow through this control volume yields
control volume
−u1ρ u1A1 + u2 ρ u2 A2 = p1A1 − FA − p2 A2
(1)

From the conservation of mass equation m = ρ1A1V 1 = ρ2 A2V 2 , we obtain

m = ρ u1A1 = ρ u2 A2
Thus Eq. (1) may be expressed as

m( u2 − u1 ) = p1A1 − FA − p2 A2
or, with p2 = 0 psi

FA = p1A1 + m( u2 − u1 ) = p1
π D12
π D2
− ρ u1 1 ( u2 − u1 )
4
4
and
lb  π ( 3 in.)

FA =  75 2 
 in. 
FA = 352 lb
4
2

ft
ft 
− 25 
2 100
2
ft  π ( 3 in.) 
s
s   lb ⋅ s 
slugs  

− 1.94 3   25 
ft   s 

4

in.2 
144


ft 2 

1
 slug ⋅ ft 


Problem 5.42
Exhaust (assumed to have the properties of standard air) leaves the 4-ft-diameter chimney
shown in the figure below with a speed of 6 ft/s . Because of the wind, after a few diameters
downstream, the exhaust flows in a horizontal direction with the speed of the wind, 15 ft/s .
Determine the horizontal component of the force that the blowing wind exerts on the
exhaust gases.
15 ft/s
15 ft/s
6 ft/s
4 ft
Solution 5.42
cs
V2 = 15 ft/s
15 ft/s
(2)
RX
p1 = p2 = 0
6 ft/s = V1
4 ft
y
(1)
x
For the control volume indicated the x-component of the momentum equation
CS uρ V ⋅ ndA =  Fx becomes
V2 ρV2 A2 = RX , where R X is the net horizontal force that the wind puts on the exhaust gases.
Thus,


RX = m2 V2 where m2 = m 1 = ρ A2V2 = ρ AV
1 1

slugs   π
slugs

2   ft 
or m2 =  0.00238
  (4 ft)   6  = 0.179
s 4
s

 s 
Hence,
RX = 0.179
slugs  ft 
slugs ⋅ ft
= 2.69 lb
15  = 2.69
s  s
s2
Problem 5.43
Air at T1 = 300 K , p1 = 303 kPa , and V1 = 0.5 m/s enters the Venturi as shown in the figure
below. The air leaves at T2 = 220 K and p2 = 101kPa ; A1 = 0.6 m 2 and A2 = 1.0 m 2 .
Calculate the horizontal force required to hold the Venturi stationary.
V1 = 0.5 m/s
V2
T1, p1
T2, p2
F
patm = 101 kPa
Solution 5.43
GIVEN: T1 = 300 K , p1 = 303 kPa , V1 = 0.5
m
, T2 = 220 K , and p2 = 101kPa , A1 = 0.6 m 2
s
and A2 = 1.0 m 2 .
FIND: Horizontal force to hold ventori stationary.
SOLUTION:
Apply the linear momentum equation in the x-direction to the control volume enclosing the
ventori.


M x,out − M x,in =  Fx
Now


M x,out = mV2 = ρ1AVV
1 1 2,


2
M x,in = mV1 = ρ1AV
1 1 ,
and
 Fx = F + p1A1 + patm ( A2 − A1 ) − p2 A2
= F + ( p1 − patm )A1 − ( p2 − patm )A2 .
The linear momentum equation is
F + ( p1 − patm ) A1 − ( p2 − patm ) A2 = ρ1AV
1 1 (V2 −V1 )
or
F = ( p2 − patm ) A2 − ( p1 − patm ) A1 + ρ1AV
1 1 (V2 −V1 )
F
A1
X
A2
Assuming the air is an ideal gas
N 

303000 2 

p
kg
m 

= 3.52 3
ρ1 = 1 =
RT1 
N⋅m 
m
 287 kg ⋅ K  ( 300 K )


V 2 is fluid from the continuity equation,
ρ1AV
1 1 = ρ2 A2V2
or
 ρ  A 
 p  RT2   A1 
 p1  T2   A1 
V2 = V1  1  1  = V1  1 
   = V1     
 ρ2  A2 
 RT1  p2   A2 
 p2  T1   A2 
m   303kPa  220 K   0.6 m 2 
m

=  0.5  
 = 0.66

 
2
s   101kPa  300 K   1.0 m 
s

Then,
3
kN
kN

 10 N
F = (101 − 101) 2 (1.0 m 2 ) − (303 − 101) 2 (0.6 m 2 ) 

 kN
m
m
kg 
m
m N ⋅ s2

2 
+  3.52 3  (0.6 m )  0.50  (0.66 − 0.5)
s 
s kg ⋅ m


m 
= −121200 N
or
F = 121.2 kN, acting to left on venturi.
Problem 5.44
Water flows steadily from a tank mounted on a cart as shown in the figure below. After the
water jet leaves the nozzle of the tank, it falls and strikes a vane attached to another cart.
The cart’s wheels are frictionless, and the fluid is inviscid. (a) Determine the speed of the
water leaving the tank, V1, and the water speed leaving the cart, V 2 . (b) Determine the
tension in rope A. (c) Determine the tension in rope B.
Nozzle area = 0.01 m2
2m
Horizontal free jets
V1
4m
V2
Rope A
Rope B
Solution 5.44
CV1
h1 = 2 m
Horizontal free jets
(1)
h2 = 4 m
CV2
Nozzle area = 0.01 m2
V1
(2)
V2
Rope A
Rope B
(a) From the equation V = 2
V1 = 2 gh1 =
γh
= 2 gh , we get
ρ
( 2 )  9.81 2  ( 2 m ) = 6.26
m

s 
m
s
(b) For CV1, the linear momentum equation leads to
CS Vρ V ⋅ ndA =  FCV
1
1
V1ρV1A1 = TA
2
m 
kg  
m

2  1N ⋅ s 
6.26
999
6.26
0.1m
= 392 N = TA





 kg ⋅ m 
s 
s 

m3  


(
)
(c) For CV2 , the linear momentum equation leads to
CS Vρ V ⋅ ndA =  FCV
2
2
V2 ρV2 A2 = TB
From the equation V = 2
(1)
γh
= 2 gh , we get
ρ
m
m

V2 = 2 g ( h2 + h1 ) = 2  9.81  ( 4 m + 2 m ) = 10.8
s 
s

Also, from conservation of mass
Q2 = Q1
or
V2 A2 = V1A1
so Eq. (1) becomes
V2 ρV1A1 = TB
2
m 
kg 
m

2  1N ⋅ s 
 = 675 N = TB
10.8   999 3  6.26  (0.01m ) 
s 
s 
m 

 kg ⋅ m 
Problem 5.45
Determine the magnitude and direction of the anchoring force needed to hold the
horizontal elbow and nozzle combination shown in the figure below in place. Atmospheric
pressure is 100kPa (abs) . The gage pressure at section (1) is 100 kPa . At section (2), the
water exits to the atmosphere.
160 mm
V2
y
Section (2)
x
300 mm
V1
Water
Section (1)
p1 = 100 kPa
V1 = 2 m/s
Solution 5.45
160 mm
V2
y
Section (2)
x
300 mm
V1
Water
Section (1)
p1 = 100 kPa
V1 = 2 m/s
The control volume shown in the sketch above is used. Application of the y-direction
component of the linear momentum equation yields
Ry = 0
Application of the x-direction linear momentum equation leads to
− u1 ρ u1A1 − u2 ρ u2 A2 = p1A1 − Rx + p2 A2
From the conservation of mass equation

m = ρ u1A1 = ρ u2 A2
Thus
Rx = ρ u1A1 ( u1 + u2 ) + p1A1 + p2 A2 = ρ u1
π D12 

π D12
+ ( 0 ) A2
 u1 + 2 u1  + p1
4 
4
D2 
D12
or




2
kg   m  π ( 300 mm )  m  ( 300 mm )  m  

2 +
2 
Rx =  999 2   2 
2 
2
m  s  4 

mm   s  
mm   s  

 1000
 
 160

m  
m 



2
+ (100 kPa )
and
Rx = 8340 N
π
( 300 mm )2  1000 N 

2 2
4
mm   m ⋅ kPa 
 1000

m 

Problem 5.46
The figure below shows a lateral pipe fitting. This particular fitting has a mainline diameter
of 4.0 in. The diameter of the lateral is 3.0 in., and the lateral angle is 45  ; 60 °F water is
flowing in the lateral. Measurements show that the pressure at point 1 is 34.0 psig , the
pressure at point 2 is 35.0 psig , the pressure at point 3 is 33.5 psig , and the flowrate at
3
point 2 is 1.0 ft s . Determine the horizontal and vertical force components (Fx and Fy)
required to hold the lateral fitting stationary. Neglect gravity. Q1=1.63ft3 s .
Q3
D3 = 3 in.
Fx
θ = 45°
3
D1 = 4 in.
Q1
Q2
1
2
y
Fy
x
Solution 5.46
First apply the continuity equation
Q1 = Q2 + Q3  Q3 = Q1 − Q2 = 1.63
ft3
ft3
ft3
− 1.0
= 0.63
s
s
s
For the “x” direction
 Fx = Fx + p1A1 − p2 A2 − p3 A3 cos 45 = ρQ2V2 + ρQ3V3 cos 45 − ρQ1V1
so

Fx = ρQ2V2 + ρQ3V3 cos 45 − ρQV
1 1 + p2 A2 + p3 A3 cos 45 − p1A1
but
A=
π
4
substituting
D2
Fx =
and V =
4 ρQ22
+
2
π D2
Q
4Q
so V =
A
π D2
4 ρQ32
4 ρQ12
π D3
π D22
cos 45 −
2
+ p2
π
4
D22 + p3
π
4
D32 cos 45 − p1
gathering terms and substituting numerical values and conversion factors gives
2
4 ρ  Q22 Q32
 Q1  π

2
2
2
Fx =
 2 + 2 cos 45 − 2  + ( p2 D2 + p3D3 cos 45 − p1D1 )
π  D2 D3
D2  4
π
4
D12
6
6
6 
slugs 
2 ft
2 ft
2 ft
1.0
0.63
1.63


ft3 
s2 +
s2 cos 45 −
s2 
Fx =
 ( 0.333ft )2 ( 0.25 ft )2
π
( 0.333ft )2 




lb
lb
lb
π

+  35.0 2 × 16 in.2 + 33.5 2 × 9 in.2 cos 45 − 34.0 2 × 16 in.2 
4
in.
in.
in.

4 × 1.94
Fx = +154.2 lb (acts to the right)
For the “y” direction
 Fy = Fy − p3A3 = ρQ3V3 sin 45 − 0  Fy = p3A3 + ρQ3V3 sin 45
The weight of the fitting and the water in it have been neglected. Substituting as before
2
2 4 ρQ3
Fy = p3 D3 +
sin 45
2
4
π D3

slugs 
ft 6
4 ×1.94 3  0.632 2

ft 
s sin 45  + π  33.5
Fy =

2
π
π
 ( 0.25ft )



× 9 in.2 sin 45 
in.

lb
 4


Fy = +28.0 lb (acts up)
2
Problem 5.47
Water flows steadily between fixed vanes, as shown in the figure below. Find the x and y
components of the water’s force on the vanes. The total volume flowrate is 100 m 3 /s,
pressure p1 = 150 kPa , and pressure p2 = 101 kPa .
V1 = 10 m/s
30°
45°
A2 = A1
y
A1
x
Solution 5.47
GIVEN: Water flows between fixed vanes as shown in the figure in the problem.
Q = 100 m 3 / s , p1 = 150 kPa , p2 = 101kPa , 20 °C water.
FIND: FX and FY on vanes by water.
30°
SOLUTION: The control volume for flow between two
vanes is shown below. Now
C.V.
45°

m3 
100


s 
Q

A1 =
=
= 14.14 m 2
m
V1 cos θ1 


 10  cos 45
 s 
(
p2, A2
y
p1, A1
)
x
θ1 = 45°
θ2 = 30°
Assuming constant fluid density and applying the continuity equation to the control
volume gives
V1 cos θ1A1 = V2 cos θ 2 A2
Since A1 = A2 = A, the continuity equation gives
 m
10  cos 45

V cos θ1  s 
m
=
= 8.165
V2 = 1

cos θ2
s
cos30
Assuming steady state, the linear momentum equation
in the x-direction for the control volume is


M x,out − M x,in =  Fx
Fy
p1A
Fx
p2 A
where

M x,out = ρ AV2 cos θ 2 (V2 cos θ 2 ) ,

M x,in = ρ AV1 cos θ1 (V1 cos θ1 ) ,
and
 Fx = Fx + p1A − p2 A
where p is taken as an absolute pressure. Fx (and FY ) is the force of the vane on the water.
The linear momentum equation is
ρ AV22 cos2 θ2 − ρ AV12 cos2 θ1 = FX + p1A − p2 A
or
Fx = ρ A(V22 cos 2 θ 2 −V12 cos 2 θ ) + ( p2 − p1 )A
Substituting values
(
2
2

 (101 − 150 ) kPa 14.14 m
kg 
m

 m
Fx =  998 3  14.14 m 2  8.165  cos 2 30 −  10  cos 2 45  −
s 


 s 
m 



 kPa 


 1000 N 
m2 

= −693000 N or Fx = 693kN acting to left on the water.
(
)
The force on the vane is
Fx = 693 kN acting to right on vanes.
The linear momentum equation in the y-direction for the control volume is


M y,out − M y,in =  Fy
where

M y,out = ρ AV2 cos θ2 (V2 sin θ2 ) ,

M y,in = ρ AV1 cos θ1 (V1 sin θ1 ) ,
and
 Fy = Fy
The linear momentum equation is
ρ AV22 cos θ2 sin θ2 − ρ AV12 cos θ1 sin θ1 = FY
or
FY = ρ A(V22 cos θ2 sin θ2 −V12 cos θ1 sin θ1 ).
2
)
Using the table below,
2
2
2


kg 
m
m


 

  N ⋅s 
2 
Fy =  998 3  14.14 m  8.165  cos30 sin30 − 10  cos 45 sin 45  

s 
m 

 s 

  kg ⋅ m 
(
)
= −298000 N = −298kN
The force on the vane is the reaction to this force or
Fy = 298kN acting up on the vanes.
Problem 5.48
The hydraulic dredge shown in the figure below is used to dredge sand from a river bottom.
Estimate the thrust needed from the propeller to hold the boat stationary. Assume the
specific gravity of the sand/water mixture is SG = 1.4.
2-ft diameter
30 ft/s
30°
9 ft
7 ft
Prop
Solution 5.48
2-ft diameter
30 ft/s
30°
9 ft
7 ft
Prop
Using the control volume shown by the broken line in the sketch above, we use the
horizontal or x-component of the linear momentum equation to get
Fx = ρ2 A2V2V2,x = ρH2O ( SG )2
π d22
4
V2V2 cos30
where section 1 is where the flow enters the control volume vertically and section 2 is where
the flow leaves the control volume at an angle of 30° from the horizontal direction. Note
that there is no horizontal direction linear momentum flow at section 1.
slugs 

Fx =  1.94 3  (1.4 )
ft 

Fx = 6650 lb
lb 

1

ft   ft 
ft ⋅ slug 


 30   30  cos 30 
 s  s 
 s2





π ( 2 ft ) 
2
4
Problem 5.49
A static thrust stand is to be designed for testing a specific jet engine, knowing the following
conditions for a typical test.
intake air velocity = 700 ft/s
exhaust gas velocity = 1640 ft/s
intake cross section area = 10 ft 2
intake static pressure = 11.4 psia
intake static temperature = 480 R
exhaust gas pressure = 0 psig
Estimate a nominal thrust to design for.
Solution 5.49
Control volume
p1A1
p2 A2
FA,x
A control volume that contains the entire engine and the fluid in the engine as indicated in
the sketch is used. Application of the horizontal or x-direction component of the linear
momentum equation leads to
−u1ρ1u1A1 + u2 ρ2 u2 A2 = p1A1 + FA,x
or
FA,x = −u1ρ1u1A1 + u2 ρ2 u2 A2 − p1A1
The conservation of mass principle yields
ρ1u1A1 = ρ 2 u2 A2
Thus
FA,x = ρ1u1A1 ( u2 − u1 ) − p1A1
or since
p
ρ1 = 1
RT1
then
FA,x =
p1
u1A1( u1 − u2 ) − p1A1
RT1


2 

lb  
ft 
ft
ft 
in.
lb 

2 

11.4 2   700  10 ft 1640 − 700  144 2  1
s
s
s 
in.  
ft   slug ⋅ ft 




s2 

=
(
FA,x
)

ft ⋅ lb 
 1716

slug ⋅ R 

lb
lb  
in.2 

−  11.4 2 − 14.7 2  144 2  10 ft 2
in.
in.  
ft 

(
and
FA,x = 17900 lb
)
( 480 R )

Problem 5.50
A vertical jet of water leaves a nozzle at a speed of 10 m/s and a diameter of 20 mm . It
suspends a plate having a mass of 1.5 kg as indicated in the figure below. What is the
vertical distance h?
h
Solution 5.50
Rz
V1
CV
h
V0
To determine the vertical distance h, we apply the vertical direction component of the linear
∂
dA =  Fcontents of the to the water in the
momentum equation
 Vρ d V +  vρ v ⋅ n
∂t CV
control volume
CS
control volume shown in the sketch above. Thus,
2
−Rz − ρ g V water = −V1ρ AV
1 1 = − ρV1
π
4
D12
(1)
The vertical reaction force of the plate on the water is equal in magnitude to the weight of
the plate, or
m

Rz = gmplate =  9.81  (1.5 kg ) = 14.7 N
s 

Also, the weight of the water within the control volume, ρ gV water , is negligible, and the
mass flowrate is

kg  π
m
kg
2

m = ρ AV
1 1 = ρ A0V0 =  999 3  ( 0.02 m )  10
 = 3.13
s
m 4

 s 
Thus, Eq. (1) becomes

−14.7 N = −V1 m
or
V1 =
14.7 N
m
= 4.70
kg
s
3.13
s
1
From the Bernoulli equation, p + ρV 2 + γ z = constant along streamline, we have
2
1
1
þ0 + ρV02 + γ z0 = þ1 + ρV12 + γ z1, where þ0 = þ1 = 0, z0 = 0, z1 = h
2
2
1
1
Thus, ρV02 = ρV12 h
2
2
or since γ = ρ g
2
1
1
2
2
2
2 m
h=
(V0 −V1 ) =
()10 − 4.70 2 = 3.97 m
m
2g

s
2  9.81 2 
s 

Problem 5.51
A horizontal, circular cross-sectional jet of air having a diameter of 6 in. strikes a conical
deflector as shown in the figure below. A horizontal anchoring force of 5 lb is required to
hold the cone in place. Estimate the nozzle flowrate in ft3/s. The magnitude of the velocity
of the air remains constant.
60°
6 in.
FA = 5 lb
Solution 5.51
Section (2)
Control volume
60°
6 in.
FA = 5 lb
Section (1)
The control volume shown in the sketch is used. Application of the axial or x-direction
component of the linear momentum equation yields
−u1ρu1A1 + u2 ρu2 A2 =− FA,x
With the conservation of mass principle, we can conclude for this in compressible flow that
u1A1 = u2 A2 = Q
Also
u2 = V cos 60
and
u1 = V =
Q
A1
Thus
−V ρQ + V cos 60 ρQ = − FA,X = −
Q2
Q2
ρ+
cos 60 ρ
A1
A1
1

 π D12   2
1
F
 A,X 
 
FA,X A1

2 
 4  
=
Q=
 
 ρ (1 − cos 60 ) 
 ρ (1 − cos 60 ) 




or
Thus
1
2







(5 lb)(π )(6 in.)2


Q=

 

2 

  0.00238 slugs  1 − cos 60 (4) 144 in.   1lb  


 
ft3 
ft 2   slug ft  




s2  


(
and
Q = 28.7
ft3
s
)
Problem 5.52
Calculate the pressure change (p2 - p1) for the jet pump shown in the figure below. The fluid
is 20 °C water. Assume negligible friction at the walls and uniform pressure over each flow
area.
Vp
D0
Dp
Vs
1
2
D0 = 1.0 m
Dp = 0.25 m
Vs = 1.07 m/s
Vp = 4.0 m/s
Solution 5.52
Assume that the two streams are fully mixed at “2.” The continuity equation gives
Q1s + Q1p = Q2
Assume uniform flow in each stream or substream
V1s As +V1 p A1 p = V2 A2  V2 =
V1s As +V1 p A1 p
A2
m π
m π
× (1.02 − 0.252 ) m 2 + 4 × (0.25)2 m 2
m
s 4
s 4
=
= 1.253
π
s
(1.02 )
4
Momentum
1.07
 Fx = p1A1 − p1A1 = ρ (V22 ) A2 − ρ (V12s ) A1s − ρ (V12p ) A1 p
or
p2 − p1 = ρ (V22 )
= 1.94
A1 p
A
A2
− ρ (V12p )
− ρ (V22s ) 2 s
A2
A2
A2
2
2
2 2
2 2
2 2
slugs 
2 m
2 m 0.25 m
2 m 1 − 0.25 m 
×
−
−
1.253
1
4
1.07
[
]


ft3 
s2
s2 12 m 2
s2
12 m 2

p1 − p2 = −0.976 Pa
or
p2 − p1 = 0.976 Pa
Problem 5.53
Air flows into the atmosphere from a nozzle and strikes a vertical plate as shown in the
figure below. A horizontal force of 12 N is required to hold the plate in place. Determine
the reading on the pressure gage. Assume the flow to be incompressible and frictionless.
p=?
12 N
Area = 0.003 m2
Area = 0.01 m2
Solution 5.53
p=?
Control
volume
(2)
9N
(1)
Area = 0.003 m2
(3)
Area = 0.01 m2
To determine the static gage pressure at station (1), we first consider the frictionless and
incompressible flow of air from (1) to (2). Bernoulli equation for this flow is
0 gage
p1 V12 p2
+
=
ρ 2
ρ
V22
+
2
(1)
We note that V1 and V 2 are linked by the continuity (conservation of mass) equation
Q1 = Q2 or AV
1 1 = A2V2
(2)
Combining Eqs. (1) and (2), we obtain
2
 A2 
 V2 
p1  A1  V22
+
=
ρ
2
2
(3)
To determine V 2 , we use the linear momentum equation for the flow from (2) to (3). For the
control volume sketched above the linear momentum principle yields
−V2 ρV2 A2 = − 9 N
or
V2 =
(12 N ) =
ρ A2
12 N
kg   N ⋅ s 2 

2
1.23
1
 0.003 m

3 

kg
m
⋅
m 


(
)
and
V2 = 57
m
s
Now, with Eq. (3)
2

 A2  
 2  V2  
A
V

p1 = ρ  2 −  1  
2
2






or

 0.003 m 2   m   
57   
  57 m  
2 

 
s   
kg   N ⋅ s 2   
s   0.01m  

p1 = 1.23 3   1
−



2
m   kg ⋅ m   2






and
p1 = 1820
N
m2
= 1820 Pa = 1.82 kPa
Problem 5.54
Water flows from a large tank into a dish as shown in the figure below. (a) If at the instant
shown, the tank and the water in it weigh W1 lb , what is the tension, T1 , in the cable
supporting the tank? (b) If at the instant shown, the dish and the water in it weigh W2 lb ,
what is the force, F2 , needed to support the dish?
T1
Tank
10 ft
0.1-ft diameter
12 ft
2 ft
Dish
F2
Solution 5.54
For part (a), we apply the vertical component of the
linear momentum equation to the contents of control
volume A, CVA , to get
−Vout ρVout Aout = T1 − W1
T1
CVA
(1)
Tank
To get value of Vout, we apply Bernoulli’s equation to
the flow from the free surface of the water in the tank to
the tank outlet to get
Vout = 2 ghA =
( 2 )  32.2 2  (10 ft ) = 25.4
ft

s 
ft
s
hA = 10 ft
0.1-ft diameter
12 ft = hB
Then from Eq. (1), we get
ft  
slugs  
ft  π (0.1ft)2

−  25.4  1.94 2   25.4 
s 
s
4

ft  
= T1 −W1
slug ⋅ ft
1
Ib ⋅ s2
and
T1 = W1 − 9.8 lb
2 ft
Dish
CVB
F2
For part (b), we apply the vertical component of the linear momentum equation to the
content of CVB to get

Vinto minto = F2 −W2
CVB
(2)
CVB
To get Vinto, we use Bernoulli’s equation between free surface water in tank to free surface
of water in dish to get
ft 
ft

Vinto = 2 g ( hA + hB ) = 2  32.2 2  (10 ft + 12 ft ) = 37.6
s
s 

CVB
•


CVB
CVB
tank
For minto ,we use from conservation of mass, minto = mout of = ρVout Aout
so from Eq. (2), we get
2
ft  
slug  
ft  π ( 0.1ft )  lb ⋅ s 2 

1
= F2 −W2
 37.6  1.94 3   25.4 


s 
s
4
ft  

 slug ⋅ ft 
and
F2 = W2 + 14.7 lb
Problem 5.55
The figure below shows the configuration of the center (tail-mounted) jet engine on an
airliner. The airliner is cruising at altitude, and the velocities shown are relative to an
observer on board. Calculate the thrust force that the engine exerts on the airplane.
D1 = 1.0 m
m• fuel = 3 kg/s
D2 = 0.75 m
Engine
V1 = 250 m/s
ρ1 = 0.4 kg/m3
p1 = pamb
F = thrust force of engine on plane
ρ2 = 0.44 kg/m3
p2 = 63 kPa
= 40 kPa
(nozzle exit only)
Solution 5.55
FIND: Trust force F of engine on airplane.
SOLUTION: Apply the linear momentum equation to a control volume enclosing the
engine and ducting. F ′ is the force of the airplane on the engine and ducting. For an
observer on the airplane and steady static,
m• fuel
patm
p1 A1
p2 A2
V1
V2
F'
+X



+  Fx = moutVx out − min Vxin




F ′ − ( p2 − patm ) A2 = m2 V2 x − m1V1x ,
or
F ′ = ( p2 − patm ) A2 + m2 V2 x − m1V1x
Now

π 2 
m1 = ρ1AV
1 1 = ρ1  D1 V1
4

kg  π
m
kg
2

=  0.4 3  (1.0 m )  250  = 78.54 ,
s 
s


m 4



kg
m2 = m1 + m fuel = (78.54 + 3.0)
s
kg
= 81.54
s
A2 =
π
4
D22 =
π
4
( 0.75 m )2 = 0.442 m 2 ,
kg 

81.54 

m2
m
s 

V2 =
=
= 419.3
kg 
s
ρ2 A2 
2
 0.44 3  0.442 m
m 

or
N
F ' = (63 − 40) ×103 2 ( 0.442 m 2 )
m

(
)
kg 
m 
kg  
m   N ⋅ s2 

+  81.54  419.3  −  78.54   250.0   

s 
s  
s 
s    kg ⋅ m 

= 24700 N.
( acts to right )
The thrust force F of the engine on the airplane is
F = 24.7 kN
acting to left
Problem 5.56
The plate shown in the figure below is 0.5 m wide perpendicular to the paper. Calculate the
velocity of the water jet required to hold the plate upright.
d = 10.0 cm
Fj
V
20 °C water jet
H = 0.5 m
Fw
20 °C water
Pivot
Solution 5.56
The forces on the plate, F j and Fw are shown in the figure. Taking moments about the pivot

F j  j = Fw  w  F j = Fw w
j
where
π

F j = ρV j2  d 2  (from the momentum equation)
4 
H
Fw = γ
Hw w = plate width (from hydrostatics)
2
2
H
and  j = H
w = H − H =
3
3
3
H
2 γ H 2w
2
m 0.52 m 2 × 0.5 m
π

w  V j2 =

V
=
×
9.81
×
substituting ρV j2  d 2  H = γ
j
6
3 ρ πd2
3
4 
s2
π × 0.12 m 2
V j = 3.24
m
s
Problem 5.57
Two water jets of equal size and speed strike each other as shown in the figure below.
Determine the speed, V , and direction, θ , of the resulting combined jet. Gravity is
negligible.
V
θ
V2 = 10 ft /s
0.1 ft
90°
0.1 ft
V1 = 10 ft /s
Solution 5.57
V
θ
y
x
V2 = 10 ft /s
0.1 ft
90°
0.1 ft
V1 = 10 ft /s
For the control volume shown in the sketch above the linear momentum equation for the x
and y directions are, for the x direction
−V2 ρV2 A2 + (V cos θ ) ρVA = 0
(1)
and for the y direction
−V1ρV1A1 + (V sin θ ) ρVA = 0
(2)
Also for conservation of mass, we have
ρ1V1A1 + ρV2 A2 − ρVA = 0
(3)
From Eqs. (1) and (2), we get
V22 A2
V12 A1
=
cos θ
= cot θ
sin θ
so
  ft 2 ( 0.1ft )2 
 10  π

2
s
4
−1 V2 A2
−1  
 = 45
= cot
θ = cot
2
2 
2

V1 A1
0.1ft )
 10 ft  π (

4
  s 

Now, combining Eqs. (2) and (3), we get
−V12 A1 +V sin θ (V1A1 + V2 A2 ) = 0
or
V=
V12 A1
sin θ (V1A1 + V2 A2 )
 ft  ( 0.1ft )
10  π
4
 s
2
V=
2
 ft 2 ( 0.1ft )2  ft 2 ( 0.1ft )2 

(sin 45 ) 10  π
+ 10  π
4
4
 s
 s 


and
V = 7.07
ft
s
Problem 5.58
The figure below shows coal being dropped from a hopper onto a conveyor belt at a
constant rate of 25 ft 3 /s . The coal has a specific gravity ranging from 1.12 to 1.50 . The belt
has a speed of 5.0 ft/s and a loaded length of 15.0 ft . Estimate the torque required to turn
the drive pulley of the conveyor belt. The drive pulley diameter is 2.0 ft . Assume that there
is no friction between the belt and the other rollers.
Coal
hopper
15.0′
A = 12 ft2
4.0′
Solution 5.58
GIVEN: Coal dropped onto conveyor belt at Q = 25 ft 2 / s , SG = 1.12 to 1.50. Belt speed =
V = 5.0 ft/s . Drive pulley diameter D = 2.0 ft. No friction between belt and rollers.
FIND: Torque Τ required to turn drive pulley.
SOLUTION: Consider an observer on the ground writing the linear momentum equation
for the fixed control volume shown on the right and in the x-direction.
Vin
Vout
C.V.F
D
x
Assuming steady state,


M x,out − M x,in =  Fx
where

2
M x,out = ρ AoutVout
,
and
 Fx = F

M x,in = ( ρ AinVin2 )(0) = 0,
Now
Aout =
Q
Vout
so the linear momentum equation is
2
F = ρ AoutVout
= ρQVout = SG ρH2OQVout .
The torque required to drive the drive pulley is
 lb ⋅ s2 
1.12 to1.50 ) 
(
slugs   ft3   ft 
 D  SG
ρH2OQVout D =
Τ=F =
 1.94 3   25 s   5 s  (2.0 ft)  ft ⋅ slug 
2
2
ft  
2





Τ = 272 to 364 ft ⋅ lb
Problem 5.59
Determine the magnitude of the horizontal component of the anchoring force required to
hold the sluice gate in place as shown in the figure below. Compare this result with the size
of the horizontal component of the anchoring force required to hold the sluice gate in place
when it is closed and the depth of water upstream is 10 ft.
10 ft
1.5 ft
4 ft/s
Solution 5.59
The control volumes of the figures below are appropriate for use in solving this problem.
Closed sluice
gate
Control volume
Open sluice
gate
Control volume
z
H
x
H
h
u
(b)
(c)
Water only
Water only
Control volume
Control volume
u1
Rx
Rx
(12_ γ H ) Hb
(
1
_ γ H Hb
2
)
Section (2)
u2
Section
(1)
(d)
Ff
(12_ γ h ) hb
(e)
When the sluice gate is closed [see Figs. (a) and (c)] application of the x direction
component of the linear momentum equation leads to
1
1
lb 
lb
Rx = γ H 2 =  62.4 3  (10 ft)2 = 3120
2
2
ft
ft 
When the sluice gate is open [see Figs. (b) and (d)] application of the x direction component
of the linear momentum equation leads to
1
1
Rx = γ H 2 − γ h2 − F f + ρ u12 H − ρ u22 h
2
2
The exit velocity u2 may be expressed in terms of the inlet velocity u , with the conservation
of mass equation as follows
u2 = u1
H
h
Thus
2
1
2 1
2
2
2 H
Rx = γ H − γ h − F f + ρ u1 H − ρ u1
h
2
2
Assuming F f is negligibly small, we obtain
1
lb 
1
lb 
Rx =  62.4 3  (10 ft)2 −  62.4 3  (1.5 ft)2
2
2
ft 
ft 




2
2
2
 1lb  
slugs   ft 
slugs   ft  (10 ft )  1lb 

+ 1.94 3   4  (10 ft) 
 − 1.94 3   4 


ft   s 
ft   s  (1.5 ft )  slug ⋅ ft 

 slug ⋅ ft  


s2 
s2 


Rx = 1290
lb
ft
Thus it takes considerably less force to hold the sluice gate in place when it is opened as
compared to when it is closed.
Problem 5.60
Water flows steadily into and out of a tank that sits on frictionless wheels as shown in the
figure below. Determine the diameter D so that the tank remains motionless if F = 0 .
d
F
D
d
Solution 5.60
(1)
d
F
h
(3)
(2) D
d
Applying the horizontal component of the linear momentum equation to the contents of
the control volume shown in the sketch, we get:
CS Vρ V ⋅ ndA =  F
or
−V1 ρV1A1 − V2 ρV2 A2 + V3 ρV3 A3 = 0
and
−V12 ρ
πd2
4
−V22 ρ
π D2
4
+V32 ρ
πd2
4
=0
Since
V2 = V3 = 2gh , we obtain
V12 d 2 = V32 d 2 −V3D2 (1)
From the conservation of mass equation, we get
Q1 = Q3 + Q3
or
V1d 2 = V2 D 2 + V3d 2
Again, since V2 = V3 = 2 gh , we get
V1d 2 = V3D2 +V3d 2 (2)
Looking at Eqs. (1) and (2) together, we conclude
If V3 < V1 eg. (1) cannot be satisfied
eg. (2) can be satisfied
If V3 > V1 eg. (1) can be satisfied
If V3 = V1
eg. (2) cannot be satisfied
eg. (1) can be satisfied with D = 0
eg. (2) can be satisfied with D = 0
So V3 = V1 and D = 0
For V3 = V1 h must be set so that
V3 = 2 gh = V1
Problem 5.61
The rocket shown in the figure below is held stationary by the horizontal force, Fx , and the
vertical force, F z . The velocity and pressure of the exhaust gas are 5000 ft/s and 20 psia at
the nozzle exit, which has a cross section area of 60 in.2 . The exhaust mass flowrate is
constant at 21lbm/s . Determine the value of the restraining force FX . Assume the exhaust
flow is essentially horizontal.
Fx
Fz
Solution 5.61
Fx
Fz
The control volume contains the rocket and the fluid within the rocket as indicated in the
sketch. Application of the x direction component of the linear momentum equation yields
because the rocket is stationary
∂ 0
u ρ d V + V1ρ1V1A1 = Fx − p1A1
∂t CV
or
FX = p1A1 + V1 ρ1V1A1
But

m = ρ1AV
1 1
Thus

Fx = p1A1 + V1 m
(
)
lb 
ft   ld 
1
 lb

Fx =  20 f2 − 14.7 f2  60 in.2 +  5000  21 m 
s 
s 
lbm ft 

in. 
 in.
⋅ 
 32.2
lbf s2 

Fx = 3580 lbf
Problem 5.62
Air discharges from a 2-in.-diameter nozzle and strikes a curved vane, which is in a vertical
plane as shown in the figure below. A stagnation tube connected to a water U-tube
manometer is located in the free air jet. Determine the horizontal component of the force
that the air jet exerts on the vane. Neglect the weight of the air and all friction.
Stagnation
tube
Air
Fixed vane
2-in. dia.
Open
Water
Free
air jet
7 in.
30°
Solution 5.62
Stagnation
tube
(1)
Air
Fixed vane
2-in. dia.
Open
Water
Rx
Free
air jet
7 in.
(2)
30°
Note that we ignore the effect of atmospheric pressure on the value of Rx in our solution
below and use gage pressures. The atmospheric pressure force may need consideration
when identifying reaction forces. For the air flowing through the control volume sketched
above, the x-direction component of the linear momentum equation is
−V1ρ airV1A1 −V2 cos 30 ρ airV2 A2 = −Rx
(1)
Application of Bernoulli’s equation for the flow from (1) to (2) yields
V2 = V1
(2)
Then, from the conservation of mass principle
AV
1 1 = A2V2
(3)
We use the Bernoulli equation again to obtain the following equation for the stagnation
tube deceleration
V12 pstag
+
=
ρair 2
ρair
p1
(4)
For the manometer, we obtain with the equation of hydrostatics
patm + hmanoγ water − hmanoγ air = Pstag
(5)
with p1 = patm , we get by combining Eqs. (4) and (5)
γ

V1 = 2hmano  water 
 ρair 
(6)
Combining Eqs. (1), (2), (3), and (6), we obtain
γ

πd2
Rx = 2hmano  water  ρair 1 (1 + cos30 )
4
 ρair 
or
lb
( 7 in.)  62.4 2  π ( 2 in.)2 1 + cos 30
ft 

Rx = 2
2
 in.   in. 
12
12


 4
 ft   ft 
and
Rx = 2.96 lb
(
)
This is force exerted by the vane on the flowing air. The force exerted by the flowing air
exerts on the vane is equal in magnitude but opposite in direction (to the right).
Problem 5.63
Water is sprayed radially outward over 180 ° as indicated in the figure below. The jet sheet is
in the horizontal plane. If the jet velocity at the nozzle exit is 20 ft/s , determine the direction
and magnitude of the resultant horizontal anchoring force required to hold the nozzle in
place.
8 in.
0.5 in.
V=
20 ft/s
Solution 5.63
Section (1)
Control volume
FA,y
8 in.
FA,x
0.5 in. = h
V=
20 ft/s
θ
Section (2)
x
The control volume includes the nozzle and water between sections (1) and (2) as indicated
in the sketch above. Application of the y direction component of the linear momentum
equation yields
 vρ V ⋅ ndA = −FA,y
CS
or
π
FA, y = − ρ  ( −V2 cos θ )(V2 )hRdθ = ρ hRV22 (sin π − sin 0)
0
and
FA, y = 0
Application of the x direction component of the linear momentum equation leads to
 uρ V ⋅ ndA = FA,x
CS
or
π
FA,x = ρ  (V2 sin θ )(V2 )hRdθ = ρ hRV22 (cos θ − cos π )
0
and
2
 ft 

0.5 in.)( 8 in.)  20  ( 2 ) 
(
slugs 
lb 

 s
FA,x = 1.94 3 
1

ft 
 in.  in. 
ft 


 slug ⋅ 2 
12 12 
s 

 ft  ft 
FA,x = 43 lb
Problem 5.64
A sheet of water of uniform thickness (h = 0.01 m ) flows from the device shown in the
figure below. The water enters vertically through the inlet pipe and exits horizontally with a
speed that varies linearly from 0 to 10 m/s along the 0.2-mlength of the slit. Determine the y
component of anchoring force necessary to hold this device stationary.
Q
x
h = 0.01 m
0.2 m
10 m/s
y
0 m/s
Solution 5.64
Q
x
FAy
Control
volume
h = 0.01 m
0.2 m
10 m/s
y
0 m/s
A control volume that contains the box portion of the device and the water in the box as
shown in the sketch above is used. Application of the y-direction component of the linear
momentum equation yields
FAy = 
0.2
Aslit
νρ V ⋅ ndA = ρ  ν 2 hdx
0
The variation of ν with x is linear or
ν = 50 x
m
s
Thus
0.2
x3
FAy = ρ  (50x ) hdx = ρ (50) h
3
0
2
0.2
2
0
or
2
( 0.2 m )3 1 N ⋅ s2 
kg  1 

FAy =  999 3  50  (0.01m)


3

m  s 
 kg ⋅ m 
and
FAy = 66.6 N
Problem 5.65
The results of a wind tunnel test to determine the drag on a body (see the figure below) are
summarized below. The upstream [section (1)] velocity is uniform at 100 ft/s . The static
pressures are given by p1 = p2 = 14.7 psia . The downstream velocity distribution, which is
symmetrical about the centerline, is given by

y
u = 100 − 30 1 −  y ≤ 3ft
3

u = 100
y > 3ft
where u is the velocity in ft/s and y is the distance on either side of the centerline in feet (see
the figure below). Assume that the body shape does not change in the direction normal to
the paper. Calculate the drag force (reaction force in x direction) exerted on the air by the
body per unit length normal to the plane of the sketch.
V2 = 100 ft/s
V1 = 100 ft/s
u
Body
3 ft
3 ft
Section (1)
Section (2)
Solution 5.65
V2 = 100 ft/s
V1 = 100 ft/s
Streamline
u
Body
Rx
h
3 ft
3 ft
Section (1)
Control volume
Section (2)
The control volume containing air only as shown in the figure is used. Application of the x
direction component of the linear momentum equation leads to
3ft
−U1ρU1A1 + 2  u ρ udy = −Rx
(1)
0
or
3ft
Rx = ρU12 h − 2 ρ
2

 y 
 100 − 30 1 − 3  dy
0
To determine the distance h, the conservation of mass equation is applied between sections
(1) and (2) as follows
3ft
ρ hU1 = 2  ρ udy
0
Thus
3ft
h=
2

 y 
100 − 30 1 −   dy


U1 0 
 3 
or

h=
( 2 )  255
ft3 

s 

= 5.1ft
ft 

100  (1ft )
s

Then from Eq. (1)


2

slugs  
ft 
lb 

Rx =  0.00238 3  100  (5.1ft ) 1
 (1ft )
s
ft  

 slug ⋅ ft 
s2 



4 

slugs 
ft
lb 

− 2  0.00238 3   21900 2  1

ft  
s   slug ⋅ ft 



s2 

Rx = 17.1lb per ft of length normal to the plane of the sketch.
Problem 5.66
A variable mesh screen produces a linear and axisymmetric velocity profile as indicated in
the figure below in the airflow through a 2-ft-diameter circular cross-sectional duct. The
static pressures upstream and downstream of the screen are 0.2 and 0.15 psi and are
uniformly distributed over the flow cross-sectional area. Neglecting the force exerted by the
duct wall on the flowing air, calculate the screen drag force.
Variable mesh screen
D = 2 ft
Section (1)
Section (2)
p1 = 0.2 psi
V1 = 100 ft/s
p2 = 0.15 psi
Solution 5.66
Control volume
D = 2 ft
Variable mesh screen
p1A1
Rx
p2A2
Section (1)
Section (2)
p1 = 0.2 psi
V1 = 100 ft/s
p2 = 0.15 psi
Application of the axial component of the linear momentum equation to the flow through
the control volume shown in the sketch leads to
R
−V1ρV1A1 +  u2 ρ u2 2π rdr = p1A1 − Rx − p2 Α2
0
or
R
2
2
π D14
π D22
r

2 π D1
Rx = ρV1
− 2πρ  umax  rdr + p1
− p2
R
4
4
4
0

The value of umax may be obtained from conservation of mass as follows
ρV1
π D12
4
R
r

= ρ   umax 2π rdr
R
0
(1)
Thus
umax =
V1D12 R
R
(2)(4)  r 2 dr
3
3
ft 
ft
= V1 = 100  = 150
2
2
s
s
0
From Eq. (1)


2
slug  
ft  π (2 ft)2 
lb 

Rx =  0.00238 3  100 
1

s
4
ft  

 slug ⋅ ft 
s2 


slug 
ft  (2 ft)
lb 

− 2π  0.00238 3   150 
1

s  16  slug ⋅ ft 
ft  



s2 

2

2
in.2 
lb  π (2 ft)2 
in.2  
lb  π (2 ft)2 

+  0.2 2 
 144 2  −  0.15 2 
144 2 
4
4
ft 
in. 
ft  
in. 



Rx = 13.3 lb
Problem 5.67
Consider unsteady flow in the constant diameter, horizontal pipe as shown in the figure
below. The velocity is uniform throughout the entire pipe, but it is a function of time:
V = u ( t )i . Use the x component of the unsteady momentum equation to determine the
pressure difference p1 − p2 . Discuss how this result is related to Fx = max .
u (t)
ℓ
D
x
ρ = density
(1)
(2)
Solution 5.67
CV
u (t)
ℓ
D
x
ρ = density
(1)
(2)
Using the control volume shown in the sketch and applying the x-component of the
unsteady linear momentum equation to the contents of this CV , we get
∂
ρ ud V +  ρ uV ⋅ n
dA =  Fx

∂t CV
CS
or
∂ π 2 
2
2
 ρ u D   − ρ u1 A1 + ρ u2 A2 = p1A1 − p2 A2 + Fx
∂t 
4

ρ u12 A1 = ρ u22 A2 assuming u1 = u2 at every instant
Fx = 0 assuming frictionless flow
Thus,
ρ
π D2
4
∂u
∂t
l ×
mass in CV
local
acceleration
and
p1 − p2 = ρ 
∂u
∂t
= (p − p )πD
1 2 4
force
2
Problem 5.68
In a laminar pipe flow that is fully developed, the axial velocity profile is parabolic. That is,
  
 −  
  
as is illustrated in the figure below. Compare the axial direction momentum flowrate
calculated with the average velocity, u, with the axial direction momentum flowrate
calculated with the nonuniform velocity distribution taken into account.
=
u
uc
Solution 5.68
The axial direction momentum flowrate based on a uniform velocity profile with u = u is
2
MFx, uniform = uρ uA = ρ u π R 2
The axial direction momentum flowrate based on the nonuniform velocity profile is
1
1
7

r

  r r
u ρ u2π r dr = ρ uc2 2π R 2 1 −      d  
0
0
R R R
MFx,nonuniform = 
R


49 2 2
MFx,nonuniform =
ρ uc π R
60
To obtain a relationship between u and uc , we use the conservation of mass equation as
follows
R
1
 r  7  r 
r
  
ρ u π R2 = ρ 2π R 2 uc  1 −      d  
0
R
R
R


 
Thus
7
u = uc
8
And
MFx, nonuniform =
16 2 2 16
ρ u π R = MFx, uniform ≈ 1.07MFx, uniform
15
15
Problem 5.70
A Pelton wheel vane directs a horizontal, circular cross-sectional jet of water symmetrically
as indicated in the figure below. The jet leaves the nozzle with a velocity of 100 ft/s.
Determine the x-direction component of anchoring force required to (a) hold the vane
stationary, (b) confine the speed of the vane to a value of 10 ft/s to the right. The fluid speed
magnitude remains constant along the vane surface.
45°
45°
y
100
ft/s
100
ft/s
10 ft/s
x
D = 1 in.
D = 1 in.
45°
45°
(a)
(b)
Solution 5.70
Control
volume
45°
Control
volume
45°
y
100
ft/s
FA
100
ft/s
x
10 ft/s
FA
D = 1 in.
D = 1 in.
45°
45°
(a)
(b)
(a) To determine the x-direction component of anchoring force required to hold
the vane stationary, we use the stationary control volume shown above and the
x-direction component of the linear momentum equation
∂
dA =  Fcontents of the
 Vρ d V +  Vρ V ⋅ n
∂t CV
control volume
CS
Thus,



FA = m(V1 +V2 cos 45 ) = ρ AV
1 1 (V1 + V2 cos 45 ) = ρ
π D12
4
V1(V1 +V2 cos 45 )
or
2
slugs 

FA = 1.94 3 
ft 

ft 




ft  
ft 
lb 
s 


100
100
cos
45
1
+


 



2
s 
s

  slug ⋅ ft 
 in. 


(4) 12 
s2 

 ft 
π (1in.) 100
and
FA = 181lb
(b) To determine the x-direction component of anchoring force requited to
confine the vane to a constant speed of 10 ft/s to the right, we use a control
volume moving to the right with a speed of 10 ft/s and the x-direction component
of the linear momentum equation for a translating control volume
 Wρ W ⋅ ndA =  Fcontents of the
control volume
CV
Thus,

FA = ρ AW
1 1 (W1 + W2 cos 45 ) = ρ
π D12
4
W1 (W1 +W2 cos 45 )
(1)
we note that
W1 = V1 − 10
ft
ft
ft
ft
= 100 − 10 = 90
s
s
s
s
Thus, Eq. (1) leads to


2


slugs  π (1in.)  ft   ft   ft 
lb


90   90  +  90  cos 45   1
FA =  1.94 3 

2
ft   in.   s   s   s 

  slug ⋅ ft 


4  12 
s2 

 ft 
or
FA = 146 lb
Problem 5.71
The thrust developed to propel the jet ski as shown in the figure below is a result of water
pumped through the vehicle and exiting as a high-speed water jet. For the conditions shown
in the figure, what flowrate is needed to produce a 300-lb thrust? Assume the inlet and
outlet jets of water are free jets.
3.5-in.-diameter
outlet jet
30°
25-in.2 inlet area
Solution 5.71
For the control volume indicated the x-component of the momentum equation
 uρ V ⋅ ndA = Fx becomes
CS
(V1 cos 30 ) ρ ( −V1 )A1 + V2 ρ ( +V2 )A2 = Rx
(1)
V1 30° = θ
Control
surface

With m = ρ AV
1 1 = ρ A2V2 , Eq. (1) becomes


Rx = m(V2 −V1 cos θ ) = ρ AV
1 1 (V2 + V1 cos 30 )
(1)
Also, AV
1 1 = A2V2 so that
AV
25 in.2
1 1
=
V1 = 2.60V1 (2)
π
2
A2
(3.5 in.)
4
By combining Eqs. (1) and (2):
2

Rx = ρ AV
1 1 2.60 − cos 30
or
1
2



300 lb
ft
 = 22.7
V1 = 
s
 1.94 slugs  25 ft 2  (2.60 − cos30 ) 

3   144
 

ft  


Thus,
ft 
ft3
 25 2 
Q = AV
ft  22.7  = 3.94
1 1 =
s
s
 144 
(2)
(1)
where we have assumed that p = 0 on the entire
control surface and that the exiting water jet is
horizontal.
V2 =
V2
Rx = 300 lb
y
x
Problem 5.72
Thrust vector control is a technique that can be used to greatly improve the
maneuverability of military fighter aircraft. It consists of using a set of vanes in the exit of a
jet engine to deflect the exhaust gases as shown in the figure below. (a) Determine the
pitching moment (the moment tending to rotate the nose of the aircraft up) about the
aircraft’s mass center (cg) for the conditions indicated in the figure. (b) By how much is the
thrust (force along the centerline of the aircraft) reduced for the case indicated compared to
normal flight when the exhaust is parallel to the centerline?
pout = 0
Vout = 1500 ft/s
cg
20 ft
Vin = 300 ft/s
pin = 0
Vane
θ = 8°
m• in = m• out = 16 slugs/s
Solution 5.72
ℓout
pout = 0
Vout = 1500 ft/s
CV
cg
20 ft
Vin = 300 ft/s
pin = 0
ℓin
•
•
min = m out = 16 slugs/s
Vane
θ = 8°
For part (a), we apply the component of the moment-of-momentum equation that is
perpendicular to the plane of the sketch of the aircraft to the content of the control volume
shown to get


 outVout sin θ mout −  inVin min = pitching moment
( 20 ft ) 1500
ft 
slugs 
ft  slugs 

o
 sin8  16
  in  300 16

s
s 
s 
s 


−
= pitching moment
slug ⋅ ft
slug ⋅ ft
1
1
lb ⋅ s2
lb ⋅ s2

66800 − 4800  in ft ⋅ lb = pitching moment
For part (b), we apply the horizontal component of the linear momentum equation to the
contents of the control volume to get


Vout cos θ m out −Vin min = Fx
where Fx is the thrust. Therefore

Fx,θ =0 − Fx,θ =8 = Vout (cos 0 − cos8 ) mout
or
ft 
slugs 


 
1500  (cos 0 − cos8 ) 16

s
s 

Fx,θ =0 − Fx,θ =8 = 
slug ⋅ ft
1
lb ⋅ s2
and
Fx,θ =0 − Fx,θ =8 = Thrust = 234 lb
Problem 5.73
The exhaust gas from the rocket as shown in the figure (a) below leaves the nozzle with a
uniform velocity parallel to the x axis. The gas is assumed to be discharged from the nozzle as
a free jet. (a) Show that the thrust that is developed is equal to ρ AV 2 , where A = π D 2 / 4 . (b)
The exhaust gas from the rocket nozzle shown in the figure (b) below is also uniform, but
rather than being directed along the x axis, it is directed along rays from point 0 as indicated.
Determine the thrust for this rocket.
m•
D
V
(a )
O
m•
θ
D
V
( b)
Solution 5.73
m•
D
CV
V
(a)
CV
O
θ
m•
D
V
(b)
For part (a), we use a control volume around the entire rocket as shown. The axial
component of the linear momentum equation applied to the contents of this CV yields
V ρVA = thrust developed = Fx
(1)
For part (b), we again use a control volume around the entire rocket application of the
linear momentum equation to the contents of this CV yields
 u ρ V ⋅ ndA = Fx X1,, X n
or
D
2
ρ  V 2 cos2 φ zπ rdr = Fx
0
and
D
2
2πρV 2  cos2 φrdr = Fx
0
Now
D
2
2 
so 2πρV 2   2 2 rdr = Fx


r2 + 2
0 r + 

cos φ =
and
   D 2


 ln    +  2 

 2 
 ln  2 
−
 = 2πρV 2  2 ln  
 = Fx

2
2 

0




D
 ln r 2 +  2  2
2πρV 2  2 


(
)
2
( )
  D 2 2 
  D 2 
  +  
   + 1
2
2

2 2
2 2


=F
πρV  ln
= πρV  ln   
x
2




1









Now
A=
π D2
4
θ  D
and tan   =
so
 2  2

θ 
ρV 2 A ln  tan2  + 1
2

tan
2θ

 =F
x
(2)
2
The effect of θ is shown on the graph below.
For θ = 0, Eq. (2) gives thrust developed = ρV A in agreement with Eq. (1). This can be
θ 
θ
θ

shown by noting that as θ → 0, tan → 0 and thus ln tan2 + 1 → tan2 since ln ( ε + 1) ≈ ε
2 
2
2

for ε → 0.
2
Let thrust ratio =
thrust for θ ≠ 0
= Rthrust
thrust for θ = 0
Thus, from Eqs. (1) and (2)

θ  
ln tan2   + 1
2 
Rthrust = 
θ 
tan2  
2
This result is plotted.
Thrust ratio
1
Thrust ratio
0.8
0.6
0.4
0.2
0
0
10
20
30
40
50
θ (°)
60
70
80
90
Problem 5.74
Where the plume goes Commercial airliners have wheel brakes very similar to those on
highway vehicles. In fact, antilock brakes now found on most new cars were first developed
for use on airplanes. However, when landing, the major braking force comes from the
engine rather than the wheel brakes. Upon touchdown, a piece of engine cowling translates
aft and blocker doors drop down, directing the engine airflow into a honeycomb structure
called a cascade. The cascade reverses the direction of the high-speed engine exhausts by
nearly 180° so that it flows forward. As predicted by the momentum equation, the exhaust
passing through this system produces a substantial braking force—the reverse thrust.
Designers must know the flow pattern of the exhaust plumes to eliminate potential
problems. For example, the plumes of hot exhaust must be kept away from parts of the
aircraft where repeated heating and cooling could cause premature fatigue. Also the plumes
must not reenter the engine inlet, blow debris from the runway in front of the engine, or
envelop the vertical tail. (See Problem 5.74.)
Air flows into the jet engine shown in the figure below at a rate of 9slugs/s and a speed of
300 ft/s . Upon landing, the engine exhaust exits through the reverse thrust mechanism with
a speed of 900 ft/s in the direction indicated. Determine the reverse thrust applied by the
engine to the airplane. Assume the inlet and exit pressures are atmospheric and that the
mass flowrate of fuel is negligible compared to the air flowrate through the engine.
V3 = 900 ft/s
(3)
4-ft diameter
(1)
30°
V1 = 300 ft/s
(2)
V2 = 900 ft/s
Solution 5.74
V3
The momentum equation (x-component),
 uρ V ⋅ ndA = Fx, for the control volume shown
(3)
V1
30°
y
Fx
CS
x
can be written as
V1ρ ( −V1 )A1 + ( −V2 cos 30 ) ρV2 A2 + ( −V3 cos 30 ) ρV3 A3 = − Fx
(1)
30°
p1 = p2 = p3 = 0
(2)
V2
or
Fx = ( ρV1A1 )V1 + ( ρV2 A2 )V2 cos 30 + ( ρV3 A3 )V3 cos 30
But from conservation of mass,

ρV1A1 = ρV2 A2 + ρV3 A3 = m = 9
slugs
s
Also, V2 = V3, so that Eq. (1) becomes

Fx = m(V1 + V2 cos 30 ) = 9
slugs 
ft
slugs ⋅ ft
 ft 
= 9170 lb
 300 + 900 cos 30
 = 9710
s 
s
s
s2
Note direction of Fx on engine and engine on airplane.
Fx
FA
On engine
On airplane
Problem 5.75
The figure below shows a sharp-edged splitter plate used to control the flow of a liquid jet
W units wide by H1 units high. Write expressions for the deflection angle θ and the force F
of the jet on the splitter plate as a function of the fluid density ρ , H1 ,W ,V , and plate
insertion h. This force F has no components parallel to the plate. Assume that the jet flow
is inviscid, that the jet width W remains constant, H 2 H 3 = H 2′ H 3′ = h ( H1 − h ) , and
constant fluid density.
V
patm
A1 = WH1
V
A3 = WH3
θ
H3′
H2′
h
A2 = WH2
patm
V
patm
Solution 5.75
GIVEN: The figure below, sharp-edged splitter splits jet of area A1 = WH 1. Fluid density ρ ,
and plate insertion h, inviscid flow and constant jet width W , no force component parallel
to plate.
 H2 
h
, constant fluid density, neglect gravity.

=
(
)
−
H
H
h
3
1


FIND: Deflection angle θ and force F as functions of ρ , H1 ,W ,V1 , and h.
V3, Q3
θ
F
V1, Q1
y
V2, Q2
x
SOLUTION: Apply the linear momentum equation in the x-direction for steady state
condition for a control volume around the flow.


↑ +  FY =  m outVY out −  minVY in
0 = ρQ3V3 sin θ + ρQ2 ( −V2 ) − ρQ1 (0) = ρ A3V32 sin θ − ρ A2V22
or
sin θ =
A2V22
H 2V22
hV22
A3V3
H3V3
( H1 − h )V32
=
2
=
2
Applying conservation of mass to the lower portion of the jet and the upper portion of the
stream gives
H 2′V1 = H 2V2 , H3′V1 = H3V3 ,
V2  H 2′V1   H3 
=

 = 1,
V3  H 2   H ′V1 
and
sin θ =
h
H1 − h
Now apply the linear momentum equation to the control volume in the y-direction. For
steady state,



+  Fx = moutVY out −  minVY in ,
−F = ρQ3V3 cos θ + ρQ2 (0) − ρQV
1 1,
F = ρ (QV
1 1 − Q3V3 cos θ ) ,
(
)
F = ρW H1V12 − H3V32 cos θ ,
or

H3V32
2
θ
cos
F = ρWH1V1 1 −
 .
 HV2
1 1


We now apply the mechanical energy equation to the upper stream. Since the flow is given
as inviscid, we get
p V2
p1 V12
+
+ 9 z1 = 3 + 3 + 9 z3
ρ
ρ
2
2
Now p1 = p2 and assume the elevation change ( z3 − z1 ) is negligible to get
V3 = V1.
Therefore,
 H

F = ρWH1V12  1 − 2 cos θ  .
 H1

Since the continuity equation for the upper stream gives
H3′V1 = H3V3 ,
 H − h H 2 − 2H h 
1
1 
2
= ρWH1V1 1 − 1
,


H1
H1 − h


or

2h 
F = ρWH1V12  1 − 1 −

H1 

1 –
h
H
H3 = H3′ = H1 − h
so
 H −h

F = ρWH1V12  1 − 1
cos θ  ,
H1


h
we get
θ
(H1 – h)2 – h2
=
H21 – 2H1h
Problem 5.76
Motorized surfboard When Bob Montgomery, a former professional surfer, started to
design his motorized surfboard (called a jet board), he discovered that there were many
engineering challenges. The idea is to provide surfing to anyone, no matter where they live,
near or far from the ocean. The rider stands on the device like a surfboard and steers it like
a surfboard by shifting his or her body weight. A new, sleek, compact 45-hp engine and
pump was designed to fit within the surfboard hull. Thrust is produced in response to the
change in linear momentum of the water stream as it enters through the inlet passage and
exits through an appropriately designed nozzle. Some of the fluid dynamic problems
associated with designing the craft included one-way valves so that water does not get into
the engine (at both the intake or exhaust ports), buoyancy, hydrodynamic lift, drag, thrust,
and hull stability. (See Problem 5.76.)
The thrust to propel the powered surfboard shown in the figure below is a result of water
pumped through the board that exits as a high-speed 2.75-in.-diameter jet. Determine the
flowrate and the velocity of the exiting jet if the thrust is to be 300 lb . Neglect the
momentum of the water entering the pump.
Solution 5.76
Fx
V2
θ
(2)
p1 = p2
(1)
y
V1
x
The x-component of the momentum equation,  u ρ V ⋅ ndA = Fx, for the control volume
CS
shown is
( −V1 cos θ ) ρ ( −V1 ) A1 + ( −V2 ) ρV2 A2 = −Fx
or
Fx = ρV22 A2 − ρV12 A1 cos θ ≈ ρV22 A2
if the momentum of the entering water is neglected.
Thus,
slugs   π  2.75 2  

300 lb = 1.94 3 V22  
ft  

ft   4  12

or
V2 = 61.2
ft
s
and
Q = A2V2 =
π  2.75
ft 
ft3
2 
=
H
61.2
2.52



4  12
s
s

Problem 5.77
Bow thrusters In the past, large ships required the use of tugboats for precise maneuvering,
especially when docking. Nowadays, most large ships (and many moderate to small ones as
well) are equipped with bow thrusters to help steer in close quarters. The units consist of a
mechanism (usually a ducted propeller mounted at right angles to the fore/aft axis of the
ship) that takes water from one side of the bow and ejects it as a water jet on the other side.
The momentum flux of this jet produces a starboard or port force on the ship for
maneuvering. Sometimes a second unit is installed in the stern. Initially used in the bows of
ferries, these versatile control devices have become popular in offshore oil servicing boats,
fishing vessels, and larger oceangoing craft. They permit unassisted maneuvering alongside
of oil rigs, vessels, loading platforms, fishing nets, and docks. They also provide precise
control at slow speeds through locks, narrow channels, and bridges, where the rudder
becomes very ineffective. (See Problem 5.77.)
The bow thruster on the boat shown in the figure below is used to turn the boat. The
thruster produces a 1-m-diameter jet of water with a velocity of 10 m/s . Determine the
force produced by the thruster. Assume that the inlet and outlet pressures are zero and that
the momentum of the water entering the thruster is negligible.
V = 10 m/s
D=1m
Solution 5.77
Fy
V2
(2)
y
x
(1)
The y-component of the momentum equation,  νρ V ⋅ ndA = Fy, for the control volume
shown is,  νρ V ⋅ n dA + V2 ρV2 A2 = F y
CS
(1)
If the momentum of the entering water is negligible the equation becomes
2
Fy = ρV22 A2 = 999
kg  m   π
kg ⋅ m
2
=
= 78.5 kN
10
(1m)
78500





m3  s   4
s2
Problem 5.78
Water flows from a two-dimensional open channel and is diverted by an inclined plate as
illustrated in the figure below. When the velocity at section (1) is 10 ft/s , what horizontal
force (per unit width) is required to hold the plate in position? At section (1), the pressure
distribution is hydrostatic, and the fluid acts as a free jet at section (2). Neglect friction.
20°
Section (1)
Plate
10 ft/s
4 ft
Section (2)
1.0 ft
Solution 5.78
FAx
20°
Section (1)
Plate
10 ft/s
4 ft
Section (2)
1.0 ft
A control volume that contains most of the plane and the water being turned by the plate as
shown in the sketch above is used. Application of the horizontal x-direction component of
the linear momentum equation yields
1
−V1ρV1A1 +V2 sin 20o ρV2 A2 = − FAx + γW h1A1
2
From conservation of mass, we obtain
A
h
V2 = 1 V1 = 1 V1
A2
h2
Thus, Eq. (1) becomes for unit width
h 
−V12 ρ h1 +  1 V1 
 h2 
2
1
sin 20 ρ h2 = −FAx + γW h12
2
or
2
h

1
FAx = γW h12 + V12 ρ h1 −  1 V1  sin 20 ρ h2
2
 h2 
(1)
Then
2
1
lb 
ft  
slugs   lb ⋅ s2 
2 
FAxX =  62.4 3  (4 ft) + 10  1.94 3  1
 (4 ft)
2
ft 
ft   slug ⋅ ft 
 s 
2
 4 ft   ft  
slugs   lb ⋅ s2 

10
sin
20
1.94
− 
 (1ft)



 1
ft3   slug ⋅ ft 

 1ft   s  
and
FAx = 213lb
Problem 5.80
A snowplow mounted on a truck clears a path 12 ft through heavy wet snow, as shown in
the figure below. The snow is 8 in. deep and its density is 10 lbm/ft 3 . The truck travels at
30 mph. The snow is discharged from the plow at an angle of 45  from the direction of
travel and 45  above the horizontal, as shown in the figure below. Estimate the force
required to push the plow.
U = 30 mph
θ = 45°
(in plane of blade)
d = 8 in.
Solution 5.80
U = 30 mph
θ = 45°
(in plane of blade)
(2)
d = 8 in.
(1)
To estimate the force required to push the snowplow, we use the control volume shown in
the sketch above and the equation  Wρ W ⋅ ndA =  Fcontents of the . We neglect the friction
control volume
CV
force between the plow and the road surface. We also neglect any force associated with the
plow deflecting air. We only consider how much force is required to turn wet snow 135 .
For the wet snow “flow,” we get from  Wρ W ⋅ ndA =  Fcontents of the
control volume
CV

Fx = m(W1 + W2 cos 45 )
Since

m = ρ AW1
We assume W2 = W1 and get
Fx = ρ AW12 (1 + cos 45 )
Then
2

ft  

5280



mi  
 lbm   8 in. 1 
 mi  

 (1 + 0.707)
10
(12
ft)
30





ft3   12 in. ft 

 hr   3600 s  



hr  

Fx =
lbm ⋅ ft 

 32.174

lb ⋅ s 2 

Fx = 8220 lb
Problem 5.81
An incompressible fluid flows outward through a blower as indicated in the figure below.
The shaft torque involved, Tshaft, is estimated with the following relationship:

Tshaft = m r2Vθ ,2

where m = mass flowrate through the blower, r2= outer radius of blower, and Vθ ,2=
tangential component of absolute fluid velocity leaving the blower. State the flow
conditions that make this formula valid.
V2
Vθ2
r2
r1
ω
Solution 5.81
Control
volume
V2
Vθ,2
r2
r1
ω
The flow conditions that make

Tshaft = m r2Vθ ,2
(1)
valid may be identified by comparing Eq. (1) with the axial component of the equation
∂
dA =  r × Fcontents of the These conditions are
( r × V ) ρ d V +  ( r × V ) ρ V ⋅ n

∂t CV
control volume
CS
(a) stationary and nondeforming control volume (see sketch above)
(b) steady-in-the-mean flow
(c) negligible shear stress torque with respect to axis of rotation
(d) Vθ ,1 = 0
(e) no torque with respect to axis of rotation due to normal stresses
(f) uniform distribution of Vθ ,2
Problem 5.82
Water at 60 °F is flowing through the 2-in. schedule 40 steel pipe shown in the figure below
at the rate of 90 gal/min . Determine the torque developed at the base where the pipe is
supported. Neglect the pipe and water weights. Steady-state conditions apply.
Horizontal
section of
pipe is 24 in.
long.
13 in.
9.5 in.
Pipe
supported
by wall
Q = 90 gal/min
Solution 5.82
+
GIVEN: Water is flowing through a 2-in., schedule 40 steel pipe
(d = 2.067 in.) at the rate of 90 gal/min . 60  F .
z
FIND: Torque developed at the base where the pipe is supported.
24″
Neglect the pipe and water weights.
SOLUTION: Applying the angular momentum equation to a control
V
volume around the pipe gives


+  T0 z =  m( r0 × V )Zz −  m( r0 × V ) z
out
in
where the z-axis is taken as the centerline of the pipe coming out of the wall. The moment
arm r0 for the inlet flow is zero and the moment arm for the outlet flow is taken as 24 in.
lbm
For 60  F water, ρ = 62.4 3 . For 2-in. schedule steel pipe inside diameter is d = 2.067 in.
ft
Then
V=
Q 4Q
=
=
A πd2
gal   ft 3 

4  90


 min   7.48 gal 
 2.067 
ft 
 12

π
2
= 516
ft
min
and

lbm  gal   ft3 
lbm

m = ρQ =  62.4 3   90
 = 751
 
min
ft   min   7.48gal 

Then


 T0 z = m ( −r0V ) − 0 = − m r0V
2
lbm   24  
ft   min   lb ⋅ sec 2 

= −  751
ft
516

 



min   12  
min   60 s   32.2 ft ⋅ lbm 

= −6.69 ft ⋅ lb
or T0 z = 6.69 ft ⋅ lb, counterclockwise looking at wall.
Problem 5.83
Five liters/second of water enters the rotor as shown in the figure below along the axis of
rotation. The cross-sectional area of each of the three nozzle exits normal to the relative
velocity is 18 mm 2 . How large is the resisting torque required to hold the rotor stationary?
How fast will the rotor spin steadily if the resisting torque is reduced to zero and (a) θ = 0°,
(b) θ = 30 ° , (c) θ = 60 ° ?
r = 0.5 m
Nozzle exit area normal to
relative velocity = 18 mm2
θ
Q = 5 l/s
Solution 5.83
To determine the torque required to hold rotor stationary, we use the moment-of

momentum torque equation Tshaft = ( − min )( ± rinVθ in ) + mout ( ±routVθ out ) to obtain

Tshaft = m routVout cos θ
(1)
We note that

m = ρQ
(2)
and
Vout =
Q
3Anozzle
(3)
exit
Combining Eqs. (1), (2), and (3), we get
Tshaft =
ρQ 2 rout cos θ
3Anozzle
(4)
exit
To determine the rotor angular velocity associated with zero shaft torque, we again use the


moment-of-momentum torque equation Tshaft = ( − min )( ±rinVθ in ) + mout ( ±routVθ out ) to obtain,
this time with rotation,

Tshaft = m rout (Wout cos θ −U out )
(5)
We note that
U out = routω
(6)
and
Wout =
Q
3Anozzle
(7)
exit
Combining Eqs. (2), (5), (6), and (7), we get


 Q

cos θ − routω 
Tshaft = ρQrout 
 3Anozzle

 exit

(8)
(a) For θ = 0  , we use Eq. (4) to get


2
2

kg  l 
mm
N




999
5
(0.5
m)(cos
0
)
1000
1



 


m   kg ⋅ m 
m3  s 



2 
s


Tshaft =
2
l 

2
 1000 3  (3) 18 mm

m 
or
(
)
Tshaft = 231N ⋅ m
From Eq. (8), we obtain for Tshaft = 0
2
mm 
 l

5  (cos 0 ) 1000


rad
Q cos θ
s
m 

ω=
=  
= 185
l 
3Anozzle rout
s
2 
3(18mm
)
1000
(0.5 m)

3
exit
m 

(b) For θ = 30  , we use Eq. (4) to get


kg   l 
mm 
N 

 

 999 3   5  (0.5 m)(cos 30 )  1000
 1
m   kg ⋅ m 
m  s 




s2 

Tshaft =
2
l 

1000
(3)(18 mm 2 )

3
m 

or
2
2
Tshaft = 200 N ⋅ m
From Eq. (8), we obtain for Tshaft = 0
2
mm 
 l
 
 5  (cos30 ) 1000

rad
s
m 

= 160
ω=  
l 
s

3(18 mm 2 ) 1000 3  (0.5 m)

m 
(c) For θ = 60  , we use Eq. (4) to get


kg   l 
mm 
N 

 
999
5
(0.5
m)(cos
60
)
1000
1



 


m   kg ⋅ m 
m3   s 



2 
s


Tshaft =
2
l 

2
 1000 3  (3)(18 mm )

m 
or
2
2
Tshaft = 116 N ⋅ m
From Eq. (8), we obtain for Tshaft = 0
2
mm 
 l
 
 5  (cos 60 ) 1000

rad
s
m 

= 92.5
ω=  
l 
s

3(18 mm 2 ) 1000 3  (0.5 m)

m 
Problem 5.84
The figure below shows a simplified sketch of a dish-washer water supply manifold. Find
the resisting torque for a water temperature of 140 F . Q = 0.25 gal/min and ω = 3 rpm .
¼-in.-diameter holes
Q
ω
Q
Q
Q
Q
Q
Q
45°
3 in. 2 in.2 in. 2 in.
Q
Solution 5.84
GIVEN: Dishwasher supply manifold with volume flowrate Q = 0.25
gal
. Steady state.
min
ω = 3 rpm . 140 F .
FIND: Resisting torque.
SOLUTION: We set up two coordinate systems: XYZ which
is fixed and r,θ , z which is coincident with XYZ and rotates
with angular velocity ω about the Z axis. Since the manifold
is “propelled” by the jets of water, there must be a resisting
torque to keep the angular velocity constant. This resisting
torque must be in the form of air resistance and bearing
friction. The angular momentum equation in the z-direction
for the control volume enclosing the manifold gives


+  Tz =  m( r × V ) z −  m( r × V ) z
out
Z, z
ω
x
θ
r
y
in
Assuming the water enters the manifold vertically and at r = 0,

 m( r × V )z = 0.
in
Since V is the absolute velocity of the exiting fluid, then for each discharge hole of flowrate
Qhole and area Ahole ,


Q 
V = rω −  
cos 45  θ
 A hole


so


Q 
 m( r × V )z =  ρQhole rr× rω −  A 
out
holes


cos 45  θ
hole

and


Q 
+  mz =  ρQhole r 2ω − r  
cos 45 ( −Z )
 A hole


holes
Assuming the flowrate is equally divided among the holes,
gal   1 

0.25

 
ft
min   8 
Q 

=
= 0.255
 
min
gal 
 A  hole π  1   ft  
 in.  
  7.48 3 
4  4   12 in.  
ft 
gal   ft3 
ft3

Qhole =  0.25
 = 0.0334

min   7.48gal 
min

Then for 140 F water,

Q 
 − ρQhole r2ω − r  A 
holes


cos 45 Z
hole

slug  
ft3   2  rev  rad  
ft 


=  − 1.90 3   0.0334
 r  3
 2π
 − r  0.255
 ( 0.707 )  Z


min    min 
rev  
min 
ft  

holes 
slug  2 
slug ⋅ ft   

r −  0.0115
=  − 1.20
 r Z
2
min 
min2  

holes 


2 32 + 52 + 72 + 92 in.2 
2


slug  
slug ⋅ ft  2 ( 3 + 5 + 7 + 9 ) in.   min  




= − 1.20
−  0.0115
Z


2
2

  60 s 
min2 
min2 

 in. 
 in. 


12 
12 


 ft 
 ft 
or
(
)
 slug ⋅ ft 2  lb ⋅ s2 
+  Tz = −0.000746 

 s2
 slug ⋅ ft 



+  Tz = −0.000746 ft ⋅ lb
or
 Tz = 0.000746 ft ⋅ lb, counterclockwise looking down on the manifold.
Problem 5.85
The hydraulic turbine shown in the figure below has a 10  C water flowrate of 36.4 m 3 /s ,
an inlet radius R1 = 1.0 m , an outlet radius R2 = 0.50 m , a blade depth (perpendicular to
paper) h = 0.50 m , and a rotational speed N = 360 rpm ; V1 = 50 m/s , V2 = 30 m/s ,
α1 = 13.4 , and α 2 = 40 . Calculate the power transferred by the fluid to the rotor, the
inlet relative velocity W1, and the direction W1 makes with the radius at the inlet.
V1 β 1
α1
W1
U1
V2
W2
α2
Blade
U2
ω
R2
R1
Solution 5.85
GIVEN: Hydraulic turbine in the figure in the problem, 10 °C water, Q = 36.4 m 3 / s ,
R1 = 1.0 m , R2 = 0.50 m , b = blade , depth = 0.50 m, N = 360 rpm , V1 = 50 m/s ,
V2 = 30 m/s , α1 = 13.4 , and α 2 = 40 .
FIND: Power transferred by water to turbine and the magnitude and direction of W1.
SOLUTION: The power is


WS = + m (U 2V2 cos α 2 −U1V1 cos α1 )

kg  
m3 
kg

m = ρQ = 1000 3   36.4
= 36400 ,

s 
s

m  
rev   2π rad   min 
m

U 2 = R2ω = 0.5 m  360
 = 18.85 ,


min   rev   60 s 
s

rev   2π rad   min 
m

U1 = R1ω = 1.0 m  360
 = 37.70 ,


min   rev   60 s 
s

so

m  m 
  N ⋅ s2 
kg  
m  m 


WS = +  36400  18.85   30  cos 40 −  37.70   50  cos13.4  

s 
s 
s 
s  
s 


  kg ⋅ m 
= 51.0 ×106

N⋅m
s
W S = 51.0 MW
The velocity W1 is found by applying the cosine law to the inlet velocity triangle.
W1 = V12 + U12 − 2VU
1 1 cos α1
2
2
m
m
 m 
 m 
=  50  +  37.70  − 2  50  37.70  cos13.4
s  
s 
s 
s 


m
s
The angle that W1 makes with the radius can be found by finding β1. The law of cosines
gives
2
2
2
−1  U1 + W1 −V1 
β1 = cos 

2U1W1


2
2
2

m 
m  m 
  37.70  + 15.94  −  50  
s  
s  
s  
= cos −1  


m 
m

2  37.70  15.94 


s 
s 



W1 = 15.94
β1 = 133.3
β1 − 90 = 43.3
so
W1 makes an angle of 43.3 with the radius.
A second way to find this angle is to use
Q = 2π R1bW1 cos( β1 − 90 )


Q

 2π R1bW 
β1 − 90 = cos −1 



m3 


 36.4

s 


−1 
= cos 

m
 2π (1.0 m)(0.50 m)(15.94 ) 
s 


β1 − 90 = 43.4
Problem 5.86
A fan (see the figure below) has a bladed rotor of 12-in. outside diameter and 5-in. inside
diameter and runs at 1725 rpm . The width of each rotor blade is 1 in. from blade inlet to
outlet. The volume flowrate is steady at 230 ft 3 /min , and the absolute velocity of the air at
blade inlet, V1 , is purely radial. The blade discharge angle is 30  measured with respect to
the tangential direction at the outside diameter of the rotor. (a) What would be a
reasonable blade inlet angle (measured with respect to the tangential direction at the inside
diameter of the rotor)? (b) Find the power required to run the fan.
Q = 230 ft3/min
30°
V1
12 in.
5 in.
1725
rpm
1 in.
Solution 5.86
Control
volume
30°
V1
Q = 230 ft3/min
12 in.
5 in.
1725
rpm
1 in.
The stationary and nondeforming control volume shown in the sketch above is used. To
determine a reasonable blade inlet angle, we assume that the blade should be tangent to the
relative velocity at the inlet. The inlet velocity triangle is sketched below.
U1
W1
V1
θ1
With the velocity triangle, we conclude that
V 
(1)
θ1 = tan −1  1 
 U1 
Now

ft3 
m2 
144 2 
 230

min 
ft 
Q
Q
ft


=
=
= 35.1
V1 =
s 
A1 2π r1h1
s

2π ( 2.5 in.)(1 in.)  60

 min 
and
rev  rad 
( 2.5 in.) 1725
 2π

ft
min 
rev 

= 37.6
U1 = r1ω =
s 
s
 in. 
12  60

 ft  min 
Thus with Eq. (1)

ft  
  35.1 s  
  = 43
θ1 = tan −1  
  37.6 ft  

 
s  

The power required, W shaft , may be obtained with the equation


  
W shaft =  − min  ( ±UinVθ in ) + mout ( ±U outVθ out )


Thus


W shaft = m 2U 2Vθ ,2
(2)

The mass flowrate, m2 , may be obtained as follows.


3 

slugs 
ft
1 

−3 slug
m2 = ρQ =  2.38 ×10 −3 3   230


 = 9.12 ×10


s
min  60
s
ft  


 min 

Also
rev  rad 

( 6 in.) 1725
 2π

ft
min 
rev 

= 90.3
U 2 = r2ω =
s 
s
 in. 
12  60

 ft  min 
The value of Vθ ,2 may be obtained by considering the velocity triangle for the flow leaving
the rotor at section (2). The relative velocity at the rotor exit is considered to be tangent to
the blade there. The rotor exit flow velocity triangle is sketched below.
U2
Wθ ,2
W2
Vθ ,2
30°
VR,2
V2
Now
Vθ ,2 = U 2 −Wθ ,2
and

ft3  
m2 
230
144



min  
ft 2 

Q
s 

2π ( 6 in.)(1 in.)  60

V
2π r2 h2
 min  = 25.4 ft
Wθ ,2 = R,2  =
=
s
tan30
tan30
tan30
Thus
Vθ ,2 = 90
ft
ft
ft
− 25.4 = 64.9
s
s
s
and from Eq. (2)




slug
ft
ft
lb
ft ⋅ lb






90.3
64.9
1
W shaft =  9.12 × 10 −3

 = 53.4



s 
s 
s   slug ⋅ ft 
s


2 
s 


Problem 5.87
Calculate the torque required to drive the pump shown in the figure below at 30 Hz and to
deliver 20 C water at 3.0 m 3 /min .
Absolute velocity, V2
b2 = 1.0 cm
θ2 = 45°
ω, T
Absolute
velocity, V1
R2 =
0.13 m
30° = θ1
b1 =
3.0 cm
R1 =
0.065 m
Impeller
Solution 5.87
GIVEN: Pump in the figure in the problem, rotates at 30 Hz and delivers water at
3.0 m 3 / min , R1 = 0.065 m, R2 = 0.13m, b1 = 3.0 cm, b2 = 1.0 cm, θ =30 , and θ2 = 45 , 20 C.
FIND: Required torque.
SOLUTION: The absolute velocity V1 is found from
Q = AV
1 1 sin θ1 = 2π R1bV
1 1 sin θ1
or

m3   min 
3.0



min   60 s 
m
Q

=
= 8.16
V1 =

2π R1bV
s
2π ( 0.065 m )( 0.03 m ) sin30
1 1 sin θ1
The absolute velocity V 2 is found from
Q = A2V2 sin θ 2 = 2π R2 b2V2 sin θ 2
or

m3   min 
3.0



min   60 s 
Q
m

V2 =
=
= 8.66

2π R2 b2V2 sin θ 2 2π (0.13 m)(0.01m)sin 45
s
We now apply the angular momentum equation in the z-direction (perpendicular to the
paper) to a control volume enclosing the pump impeller.


+  Tz =  m( r × V ) z −  m( r × V ) z
out
in
where
Tz = T = torque on pump impeller,
π


 m ( r × V )z = ρQR2V2 sin  2 − θ2  ,
out
and
π


 m ( r × V )z = ρQRV
1 1 sin  − θ1  .
2

in
The angular momentum equation is then
π

π

T = ρQR2V2 sin  − θ2  − ρQRV
1 1 sin  − θ1 
2

2

= ρQ ( R2V2 cos θ2 − RV
1 1 cos θ1 ) .
For 20 C water,

m


0.13 m )  8.66  cos 45 
(

 N ⋅ s2 
s 
kg  
m  min  



T =  998 3   3.0




min   60 s  
m

m  

   kg ⋅ m 
 − ( 0.065 m )  8.16  cos 30 
s 



3 
T = 16.8 N ⋅ m counterclockwise, looking down on paper.
Problem 5.88
An axial flow turbomachine rotor involves the upstream (1) and downstream (2) velocity
triangles shown in the figure below. Is this turbomachine, a turbine or a fan? Sketch an
appropriate blade section and determine energy transferred per unit mass of fluid.
W1 = W2
W1
1
W2
U1 = 30 ft/s
V1 = 20 ft/s
U2 = 30 ft/s
60°
Solution 5.88
We can determine whether the axial flow turbomachine involved is a turbine or a fan by
comparing the direction of the lift force on the rotor blade section with the direction of the
blade velocity, U . If the lift force and the blade velocity are in the same direction, a turbine
is involved. If the lift force and blade velocity are in opposite directions, a fan is involved.
The direction of the lift force can be inferred from the shape of the rotor blade section
sketched to be tangent to the relative flows entering and leaving the rotor row.
The entering relative flow angle, β1, is
 ft 
 30 
s
−1 U1
−1 
β1 = tan
= tan
= 56.3
ft
V1


 20 
s


Thus, the rotor blade sections sketched below are appropriate.
60°
Lift
force
56.3°
U
Since the lift force acting on each rotor blade section is in the same direction as the blade
velocity, we conclude that this turbomachine is a turbine. The energy transferred per unit
mass is the shaft work per unit mass, wshaft , which we can determine with the equation
wshaft = −( ±U inVθ in ) + ( ±U outVθ out ) . Thus
wshaft = −U2Vθ ,2
(1)
From the velocity triangles, we obtain
Vθ ,2 = W2 sin 60 −U 2
and
W2 = W1 = V12 + U12
Thus
wshaft = −U 2
( V +U sin 60 −U )
2
1
2
1

2


2
2



ft
ft
ft
ft
1lb

 
 

wshaft = −  30    20  +  30  sin 60 − 30  

s   slug ⋅ ft 
 s   s   s 



s2 

ft ⋅ lb
wshaft = −36.8
slug
or
ft ⋅ lb
−36.8
ft ⋅ lb
slug
wshaft =
= −1.14
lbm

lbm 
 32.2 slug 


Problem 5.89
An inward flow radial turbine (see the figure below) involves a nozzle angle, α1, of 60° and
an inlet rotor tip speed, U 1, of 6 m/s. The ratio of rotor inlet to outlet diameters is 1.8. The
absolute velocity leaving the rotor at section (2) is radial with a magnitude of 12 m/s.
Determine the energy transfer per unit mass of fluid flowing through this turbine if the fluid
is (a) air, (b) water.
Section (1)
α1
r1
r2
V2 =
12 m/s
Section (2)
Solution 5.89
Section (1)
α1
r1
r2
V2 =
12 m/s
Section (2)
Fixed
control
volume
To determine the energy transfer per unit mass we use the moment-of-momentum energy
equation wshaft = −( ±U inVθ in ) + ( ±U outVθ out ) to obtain
wshaft = U1Vθ ,1
net out
(1)
and we note that the result is independent of the fluid involved. The value of Vθ ,1 can be
ascertained with the help of the section (1) velocity triangle sketched below.
V1
60° VR,1
Vθ,1
From the velocity triangle, we note that
Vθ ,1 = VR ,1 tan 60 
(2)
With conservation of mass, we obtain
VR,1 = VR,2
A2
r  m  1 
m
= VR,2 2 =  12  
 = 6.7
s
A1
r1  s   1.8 
Thus with Eq. (2), we obtain
m
m

Vθ ,1 =  6.7  tan 60 = 11.6
s 
s

and with Eq. (1), we get



m 
m
N 
N⋅m

wshaft =  6.7  11.6   1
 = 77.7
m
s 
s   kg ⋅ 
kg
net out 

2 
s 

for both (a) air and (b) water. Although the work per unit mass is the same for both air and
water, there will be a difference in total power due to the higher density of a water flow.
Problem 5.90
A sketch of the arithmetic mean radius blade sections of an axial-flow water turbine stage is
shown in the figure below. The rotor speed is 1000 rpm . (a) Sketch and label velocity
triangles for the flow entering and leaving the rotor row. Use V for absolute velocity, W for
relative velocity, and U for blade velocity. Assume flow enters and leaves each blade row at
the blade angles as shown. (b) Calculate the work per unit mass delivered at the shaft.
rm = 6 in.
1000
rpm
Rotor
Stator
U
70°
U
45°
Blade sections
at the arithmetic
mean radius
45°
Solution 5.90
rm = 6 in.
1000
rpm
Control volume
U
Control volume
U
Blade sections
at the arithmetic
mean radius
70°
45°
45°
The velocity triangles for the flow entering and the flow leaving the rotor row at the
arithmetic mean radius are sketched below.
Vx,2
U1
V1
V2
α2
W1
70°
Vθ ,1
45°
45°
Vθ ,2
U2
W2
Vx,1
At the arithmetic mean radius, the blade velocity, U , is
U1 = U 2 = rmω =
rev  rad 
 2π

ft
min 
rev 
= 52.3
s 
s
 in. 
12  60

 ft  min 
( 6 in.)1000
With the velocity triangle for the flow entering the rotor, we conclude that
V1 sin70 = Vθ ,1
(1)
V1 cos70 = VX ,1
(2)
W1 sin 45 = Vθ ,1 −U (3)
W1 cos 45 = VX ,1
(4)
From the ratio of Eqs. (3) and (4), we obtain
tan 45 =
Vθ ,1 − U
VX ,1
which when combined with Eqs. (1) and (2) yields
tan 45 =
V1 sin 70 − U
V1 cos 70
or
ft
s
V1 =
=
sin 70 − (cos 70 )(tan 45 )  sin 70 − (cos 70 )(tan 45 ) 

 

U
V1 = 87.6
ft
s
52.3
Then
ft 
ft

Vθ ,1 = V1 sin 70  =  87.6  sin 70  = 82.3
s
s

ft 
ft

VX ,1 = V1 cos 70  =  87.6  co s 70  = 29.9
s
s

and
ft 

 29.9 
ft
s
=
= 42.4
W1 =
o

s
cos 45
cos 45
VX ,1
With the velocity triangle for the flow leaving the rotor, we conclude that
W2 cos 45 = VX ,2
(5)
Vθ ,2 = U 2 − W2 sin 45
V2 sin α 2 = Vθ ,2
(7)
V2 cos α 2 = VX ,2
(8)
(6)
From the conservation of mass equation
VX ,1 = VX ,2 = 29.9
ft
s
Thus from Eq. (5)
ft 

29.9 

VX ,2
ft
s

42.4
=
=
W2 =
s
cos 45
cos 45
and from Eq. (6)
Vθ ,2 = U 2 −W2 sin 45 = 52.3
ft 
ft 
ft
−  42.4  sin 45 = 22.4
s 
s
s
The ratio of Eqs. (7) and (8) yields

ft  
22.4  


V 
s
 = 37
α 2 = tan −1  θ ,2  = tan −1  
 VX ,2 
ft
  29.9  



 
s  
and from Eq. (7)
ft 

22.4 

Vθ ,2
ft
s
=
= 37.2
V2 =

sin α 2
s
sin37
We can use the equation wshaft = − ( ±U inVθ in ) + ( ±U outVθ out ) to calculate the work per unit
mass delivered at the shaft. Thus
wshaft = −U1Vθ ,1 + U 2Vθ ,2


 
ft 
ft  
ft 
ft   
lb 
wshaft =  −  52.3  82.3  +  52.3  22.4   1

s 
s 
s 
s    slug ⋅ ft 
 


s2 

ft ⋅ lb
wshaft = −3130
slug
Problem 5.92
An incompressible fluid flows along a 0.20-m -diameter pipe with a uniform velocity of
3 m/s . If the pressure drop between the upstream and downstream sections of the pipe is
20 kPa , determine the power transferred to the fluid due to fluid normal stresses.
Solution 5.92
0.20 m
V1 = 3 m/s
V2 = 3 m/s
(1)
V1
(2)
p1 – p2 = 20kpa
p2V2
n̂
(3)
nˆ 2
nˆ 1
p1 (1)
(2)

From the equation, W normal =  σ V ⋅ ndA =  − pV ⋅ ndA,
stress

W
normal
stress
CS
CS
= −  pV ⋅ ndA = −  pV ⋅ ndA −  pV ⋅ ndA −  pV ⋅ ndA
CS
(1)
(2)
(3 )
= − p1 ( −V1 )A1 − p2 ( −V2 )A2 − p3 (0)A3
= p1V1A1 − p2V2 A2
but
m
π
, A1 = A2 = (0.20 m)2 = 0.0314 m 2
s
4
and p1 − p2 = 20 kPa
V1 = V2 = 3
Thus,
m3
 m
2
p
p
V
A
(
)
(20
kPa)
3
(0.0314
m
)
1.88kPa
=
−
=
=
W
1
2 1 1


normal
s
 s 
stress

= 1.88 × 103
N m3
N⋅m
= 1.88 × 103
2 s
s
m
Problem 5.93
A horizontal Venturi flow meter consists of a converging–diverging conduit as indicated
in the figure below. The diameters of cross sections (1) and (2) are 6 and 4 in. The velocity
and static pressure are uniformly distributed at cross sections (1) and (2). Determine
the volume flowrate ( ft 3 /s ) through the meter if p1 − p2 = 3 psi , the flowing fluid is
3
oil ( p = 56 lbm/ft ), and the loss per unit mass from (1) to (2) is negligibly small.
D1 = 6 in.
Section (1)
D2 = 4 in.
Section (2)
Solution 5.93
D1 = 6 in.
Section (1)
Control volume
D2 = 4 in.
Section (2)
The control volume shown in the sketch above is used. Application of the conservation of
mass equation to the incompressible steady flow through this control volume leads to
(1)
Q = AV
1 1 = A2V2
2
Vout
p
V2
+ gzout = in + in + gzin − loss
ρ
ρ
2
2
to the flow through this control volume yields
p2 V22 p1 V12
+
= +
(2)
ρ
2
ρ 2
Combining Eqs. (1) and (2), we obtain
p2 Q 2
p
Q2
or
+ 2 = 1+ 2
ρ A2 2 ρ A1 2
Application of the energy equation
pout
+
1
2
1
2





1
 p −p 
 
Q = 2  1 2  
  ρ   1 − 1 
 A2 A2  

1 
 2











1
  p1 − p2  
 
= 2 

1
1

  ρ 
−
2
2



 π D22   π D12   



  4   4   

 
 

1


 2












in.2  
lbm ft  
  lb  

  3 in.2   144 ft 2   32.2 lb ⋅ s2  

1




 
Q = 2
1
1
 lbm 



−
56 3 

2
2 


ft 



 
 




2  

(4 in.)2 
(
in.)
6

π
 
 π


2
2

 in. 
 in.    





 4 12 ft    
  4 12 ft  


  



Q = 2.17
ft3
s
Problem 5.94
The figure below shows the mixing of two streams. The shear stress between each fluid and
its adjacent walls is negligible. Why cannot Bernoulli’s equation be applied between points
in stream 1 and the mixed stream or between points in stream 2 and the mixed stream?
p1 = 50 psia
p3
5 ft/s
2 ft2 p1 = 62.4 lbm/ft3
2 ft2 p2 = 62.4 lbm/ft3
10 ft/s
p2 = 50 psia
Solution 5.94
GIVEN: The figure in the problem with negligible shear stress between fluids and wall.
FIND: Why Bernoulli’s equation cannot be applied between stream 1 and the mixed stream
or between stream 2 and the mixed stream.
SOLUTION: The mixing of two streams requires turbulence and/or viscous action and
Bernoulli's equation does apply to a viscous or turbulent flow.
It only applies to a ideal flow without friction, work, or heat transfer.
Problem 5.95
Liquid water at 40  F flows down a vertical, thermally insulated, 2-in. schedule 40 steel
pipe. The temperature change of the water is related to its internal energy change by
 Btu 
u2 − u1 = 32.2 
 
 (T2 − T1 ) .
 slug ⋅ F 
What is the temperature change of the water per 100 ft of drop if the pressure drop ( p1 − p2 )
per 100 ft is 12.0 psi ?
Solution 5.95
GIVEN: Liquid water at 40  F flows down a vertical, thermally insulated, 2-in. steel pipe.
∨
∨
u2 − u1 = 32.2
Btu

slug ⋅ F
(T2 − T1 ) , p1 − p2 =
12 psi
.
100 ft
FIND: Temperature change of water per 100 ft of pipe.
SOLUTION: Assume constant fluid density and apply the mechanical energy equation
from 1 to 2 gives (1 represents the higher elevation).
p1 V12
p V2
+
+ gz1 − wS = 2 + 2 + gz2 + ghL
ρ
2
ρ
2
Now wS = 0 and the continuity equation gives V1 = V2 .
Then
ghL =
p1− p2
ρ
+ g ( z1 − z2 ).
(1)
and
∨
∨
u 2 − u1 = q + ghL
(2)
The problem statement gives q = 0 so Eqs. (1) and (2) give
∨
∨
u 2 − u1 =
p1− p2
ρ
+ g ( z1 − z2 )
or
 p1 − p2

 ρ + g ( z1 − z2 ) 

T2 − T1 = 

Btu 
 32.2

slug ⋅ F 

From property table ρ = 1.938
slug
ft3
so


lb  
in.2 
 12 2   144 2 

 lb ⋅ s2  
ft  
ft 
  in.  
+  32.2 2  (100 ft) 

 ft ⋅ slug  
slug 

s





1.938 3 
ft 



T2 − T1 = 

Btu  
ft ⋅ lb 
778
 32.2



Btu 
slug ⋅ F  

T2 − T1 = 0.16 F .
Problem 5.96
A simplified schematic drawing of the carburetor of a gasoline (S = 0.75 ) engine is shown in
the figure below. The throat area is 0.5 in.2 . The running engine draws air downward
through the carburetor Venturi and maintains a throat pressure of 14.3 psia . The low
throat pressure draws fuel from the float chamber and into the air stream. The energy losses
in the 0.07-in.-diameter fuel metering line and valve are given by
KV 2
,
ghL =
2g
where K = 6.0 and V is the fuel velocity in the metering line. Assume that the air is an ideal
fluid having a constant density ρA = 0.075 lbm/ft3. The atmospheric pressure is 14.7 psia .


Calculate the air-to-fuel ratio (ma m f ).
Air
Air vent
Fuel
0.5 in.2
Float chamber
Solution 5.96
GIVEN: Carburetor shown in the figure in the problem. Throat pressures = 14.3psia ,
lbm
atmospheric pressures = 14.7 psia , constant air density ρA = 0.075 3 . Fuel specific
ft
V2
gravity = SG = 0.75. 0.07-in. diameter fuel line has energy loss given by ghL = 6.0
2
FIND: Air/fuel ratio.
SOLUTION: We first find the air flowrate. Assume in viscid flow and apply Bernoulli’s
equation to a streamline from the atmosphere (0) to the throat (1).
p0
ρair
+
V02
p
V2
+ gz0 = 1 + 1 + gz1
2
ρair 2
Assume V1 ≈ 0 and z1 ≈ z2 . This gives

lb  
in.2  
ft ⋅ lbm  
0.4
144
32.2






2 ( p0 − p1 )
in.2  
ft 2  
lb ⋅ s 2  

V1 =
= 2

lbm 
ρair

0.075




ft 2 



ft
= 222
s
The air mass flowrate is
lbm 
ft   ft 2 

2 
mair = ρair AV1 =  0.075 3  0.5 in.  222  

s   144 in.2 

ft 


(
)
lbm
s
The fuel flowrate is found by applying the mechanical energy equation from the float
chamber free surface (2) to the throat (3). For no mechanical work,
= 0.0578
p2
ρf
+
p V2
V22
+ gz2 = 3 + 3 + gz3 + ghL .
ρf
2
2
Assume z2 ≈ z3 and V2 ≈ 0 . Using the expression for ghL gives
V32
V32 patm − p1
+6
=
2
2
ρf
or
V3 =
patm − p1
.
3.5 ρ f
The numerical values give


ρ f = SG ρH O = 0.75  62.4
2
lbm 
lbm
= 46.8 3
3 
ft 
ft
and
lb  
in.2  
ft ⋅ lbm 

 0.4 2  144 2   32.2

in.  
ft  
lb ⋅ s2 

ft
= 3.36
V3 =
lbm 
s

3.5  46.8 3 
ft 

Then
lbm  π 
ft 

2
 46.8 3   4  ( 0.07 in.)  3.36 s 


ft   
m fuel = ρ f AV3 = 
2

in. 
144 2 
ft 


lbm
.
s
The air-to-fuel ratio (A/F) is
= 0.00420
lbm
0.0578
A
s
=
lbm
F 0.00420
s
A
= 13.8
F
Problem 5.97
Oil (10 psi ) flows downward through a vertical pipe contraction as shown in the figure
below. If the mercury manometer reading, h, is 100 mm, determine the volume flowrate for
frictionless flow. Is the actual flowrate more or less than the frictionless value? Explain.
300 mm
0.6 m
h
100 mm
Solution 5.97
300 mm
Section (1)
60 m
Control
volume
Section (2)
h
100 mm
The volume flowrate may be obtained with
Q = AV
1 1 = A2V2 =
π D12V1
4
=
π D22V2
4
(1)
To determine either V1 or V2 , we apply the energy equation
pout
+
ρ
Thus,
p2
ρ
+
2
Vout
p
V2
+ gzout = in + in + gzin + wshaft − loss to the flow between sections (1) and (2).
2
2
ρ
net in
V22
p V2
o
neglect
+ gz2 = 1 + 1 + gz1 + ω shaft − loss
2
2
ρ
net in
(2)
Combining Eqs. (1) and (2), we obtain
4
V22   D2   p1 − p2
1 − 
+ g ( z1 − z2 )
 =
ρ
2   D1  


p1 − p2
To determine
p1 − p2
ρ
ρ
(3)
, we use the manometer equation to obtain
 SGHg 
= gh 
− 1 − g ( z1 − z2 )
SG
oil


(4)
Combining Eqs. (3) and (4), we get
 SGHg

− 1
2 gh 
 SGoil

V2 =
4
D 
1−  2 
 D1 
or
V2 =
( 2 )  9.81 2  ( 0.1m ) 
m
s 

13.6 
− 1
 0.9

 100 mm 
1− 

 300 mm 
4
= 5.29
m
s
And from Eq. (1), we have
π ( 0.1m ) 
2
Q=
4
m
m3
 5.29  = 0.042
s 
s

Actual flowrate would be less than the frictionless value because the loss would be greater
that the zero amount used above.
Problem 5.98
An incompressible liquid flows steadily along the pipe shown in the figure below.
Determine the direction of flow and the head loss over the 6-m length of pipe.
0.75 m
1.0 m
3m
6m
1.5 m
Solution 5.98
0.75 m
(2)
1.0 m
3m
1.5 m
6m
(1)
Assume flow from (1) to (2) and use the energy equation
2
pout Vout
p
V2
+
+ zout = in + in + zin + hs − hL to get for the contents of the control volume
γ
2g
γ
2g
show:
p2
γ
+
0
V22
p V2
+ z2 = 1 + 1 + z1 + h − h
s
γ 2g
2g
Thus
h =
p1
γ
−
p2
γ
+ z1 − z2 = 3 m − 1.0 m − 1.5 m = 0.5 m
and since h > 0 , the assumed direction of flow is correct.
The flow is uphill.
Problem 5.99
A siphon is used to draw water at 70 °F from a large container as indicated in the figure
below. The inside diameter of the siphon line is 1in. and the pipe centerline rises 3 ft above
the essentially constant water level in the tank. Show that by varying the length of the
siphon below the water level, h , the rate of flow through the siphon can be changed.
Assuming frictionless flow, determine the maximum flowrate possible through the siphon.
The limiting condition is the occurrence of cavitation in the siphon. Will the actual
maximum flow be more or less than the frictionless value? Explain.
3 ft
h
1 in.
Solution 5.99
C
The flowrate, Q , can determined with
Q = ABVB =
π DB2
4
3 ft
VB (1)
A
To obtain VB , we apply the energy equation
2
Vout
pin Vin2
+
+ gzout =
+
+ gzin + wshaft − loss
2
2
ρ
ρ
net in
h
pout
1 in.
between points A and B in the sketch above to obtain
pB
ρ
0
p
V2
+ B + gzB = A
ρ
2
0
2
+ VA
2
0
+ gzA + w
0
shaft net in
− loss
B
(2)
or
VB2
= g ( zA − zB ) − loss
2
and
VB = 2  g ( h ) − loss 
(3)
With Eq. (3), we conclude that as h varies, so does VB and thus Q . For no loss, the
maximum flow will occur when the pressure at point C is just equal to the vapor pressure
of water at 0 °C.
We apply the energy equation (see above) between points A and C to get
p
VC2
+
+ gzC = A
ρ
ρ
2
pC
0
2
+ VA
2
0
0
+ gzA + wshaft net in − loss
0
(4)
Using absolute instead of gage pressures we obtain with Eq. (4)
Vc = 2 g ( zA − zC ) +
pA − pC
ρ
or
N
N 

101000 2 − 1228 2 

m
m


m
m  = 9.048 m
VC = 2  9.81 2  ( −3ft )  0.3048  + 
ft 
s


s 




kg 
N


 999.7 2  1
m   kg ⋅ m 



s2 

Since
Q = ACVC =
π DC2
4
VC
we have for the maximum flowrate through the siphon,
Q=
π (1in.)
2
2
3
m 
m

−3 m
=
×
0.3048
9.048
4.58
10

 

ft  
s 
s

in.2  
4 144 2 

ft 

With Eqs. (3) and (4), we conclude that any loss would act to lower the value of V in the
siphon and thus make the actual maximum flowrate with friction less than the maximum
flowrate without friction.
Problem 5.100
A water siphon having a constant inside diameter of 3in. is arranged as shown in the figure
below. If the friction loss between A and B is 0.8V 2 /2 , where V is the velocity of flow in
the siphon, determine the flowrate involved.
4 ft
A
12 ft
4 ft
3 in.
B
Solution 5.100
To determine the flowrate, Q , we use
Q = AV =
π D2
4
V
(1)
To obtain V , we apply the energy equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss
2
2
ρ
ρ
net in
between points A and B in the sketch. Thus,
pB
0
ρ
0
0
2
p
V2
+ B + gzB = A + VA + gzA + wshaft − loss
ρ
2
2
net in
or
V2
V2
+ gzB = gzA − 0.8
2
2
Thus
V=
ft 

32.2 2  ( 8 ft )

g ( zA − zB )
ft
s 
= 
= 16.9
0.9
0.9
s
and with Eq. (1)
2
Q=
ft 

ft3
s

= 0.830
s

in.2 
4  144 2 

ft 

π ( 3 in.) 16.9
Problem 5.101
The figure below shows a test rig for evaluating the loss coefficient, K , for a valve.
Mechanical energy losses in valves are modeled by the equation:
V 2 
,
ghL = K 
 2 


where ghL is the mechanical energy loss and V is the flow velocity entering the valve. In a
particular test, the pressure gage reads 40kPa , gage, and the 1.5-m3 catch tank fills in
2 min 55s . Calculate the loss coefficient for a water temperature of 20 °C
D1 = 6 cm
D2 = 4 cm
1
2
Valve
Pressure
gage
Catch tank
Solution 5.101
Apply the mechanical energy equation from point 1 to point 2. Assume constant water
density, steady flow in pipe, zero elevation change, and uniform velocity over each flow
area.
p1 V12 p2 V22
+
=
+
+ ghL + loss
2
2
ρ
ρ
Using p2 = 0 kPa gage,
V12 p1 1 2
= + (V1 −V22 )
ρ 2
2
and
K
K=
 V22 
+
1 −
.
ρV12  V12 
2 p1
Using V =
K=
4Q
π D2
gives
π 2 D14 p1  D14 
+ 1 −
.
8Q 2 ρ  D24 
The numerical value give
Q=
1.5 m3
m3
= 0.00857
,
175s
s
4
π 2 ( 6 cm )  40 ×103
N   kg ⋅ m 


m 2   N ⋅ s2 
 (6 cm)4 
+ 1 −
K=
,
4

4 
3 2
(4
cm)



m
kg 
cm 

8  0.00857
  998 3 100


s  
m 
m 


or
K = 4.66
Problem 5.102
For the 180° elbow and nozzle flow shown in the figure below, determine the loss in
available energy from sections (1) to (2). How much additional available energy is lost from
section (2) to where the water comes to rest?
6 in.
y
Section (2)
x
12 in.
p1 = 15 psi
V1 = 5 ft/s
Section (1)
Solution 5.102
Control
volume
6 in.
y
Section (2)
x
12 in.
p1 = 15 psi
V1 = 5 ft/s
Section (1)
For solving the first part of this problem, the control volume shown in the sketch above is
used. To determine the loss accompanying flow from sections (1) to (2), the equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin − loss can be used as follows.
ρ
ρ
2
2
1loss2 =
p1 − p2
ρ
+
V12 −V22
+ g ( z1 − z2 )
2
Since x–y coordinates are specified, we assume that the flow is horizontal and z1 − z2 = 0 .
Also, p2 = patm = 0 psi. From the conservation of mass principle, we conclude that
V2 = V1
 D2 
A1
= V1  12 
D 
A2
 2
Thus
2
4


p1 V12   D12   p1 V12   D1  
+
1− 
1− 
 = +

1loss2 =
ρ 2   D22   ρ 2   D2  




or
lb  
in.2   ft 2



15
144

5  

4
2 
2 



ft   s 
 in.  
 12 in. 
lb 


loss
1
1
=
+
−




1
2
slugs 
2   6 in.    slug ⋅ ft 




1.94 3 
s2 

ft 

ft ⋅ lb
1loss2 = 926
slug
For the second part of this problem, we consider the flow of a fluid particle from section (2)
p
V2
p
V2
to a state of rest, the equation out + out + gzout = in + in + gzin − loss leads to
ρ
ρ
2
2
V22
2
Note that we have assumed that p2 = pa = patm and z2 = za .
Thus
2 lossa =



lb 
 ft   12 in. 

5  
 1
 s   6 in.   slug ⋅ ft 
2
4


V22 V12  D12  V12  D1 
s2 

loss
=
=
=
=




2
a
2
2  D22 
2  D2 
2
2
2 lossa = 200
ft ⋅ lb
slug
4
Problem 5.103
An automobile engine will work best when the back pressure at the interface of the exhaust
manifold and the engine block is minimized. Show how reduction of losses in the exhaust
manifold, piping, and muffler will also reduce the back pressure. How could losses in the
exhaust system be reduced? What primarily limits the minimization of exhaust system
losses?
Solution 5.103
2
Vout
Vin2
We apply the energy equation pout + ρ
+ γ zout = pin + ρ
+ γ zin + ρ wshaft − ρ ( loss ) to
2
2
net in
the flow from the engine block, exhaust manifold interface to the exhaust system exit to get
2
Vout
Vin2
pin = pout + ρ
−ρ
+ ρ (loss ) (1)
2
2
With Eq. (1), we see that reduction of loss in the exhaust system result in a lower value of
pin and thus the engine back pressure. Losses in the exhaust system could be reduced by
eliminating a major loss component such as the catalytic converter and the muffler as is
often done in race cars. However, noise and emissions legislation limits the extent to which
this kind of loss reduction can also occur in conventional vehicles. Some loss reduction can
also occur by configuring the exhaust system piping with fewer bends and appropriate area
distributions. However, requirements often lead to bends and turns in the piping and costs
limit the extent of optimizing area distributions.
Problem 5.104
Smart shocks Vehicle shock absorbers are dampers used to provide a smooth, controllable
ride. When going over a bump, the relative motion between the tires and the vehicle body
displaces a piston in the shock and forces a viscous fluid through a small orifice or channel.
The viscosity of the fluid produces a head loss that dissipates energy to dampen the vertical
motion. Current shocks use a fluid with fixed viscosity. However, recent technology has
been developed that uses a synthetic oil with millions of tiny iron balls suspended in it.
These tiny balls react to a magnetic field generated by an electric coil on the shock piston in
a manner that changes the fluid viscosity, going anywhere from essentially no damping to a
solid almost instantly. A computer adjusts the current to the coil to select the proper
viscosity for the given conditions (i.e., wheel speed, vehicle speed, steering-wheel angle,
lateral acceleration, brake application, and temperature). The goal of these adjustments is
an optimally tuned shock that keeps the vehicle on a smooth, even keel while maximizing
the contact of the tires with the pavement for any road conditions. (See Problem 5.104.)
A 200-lb force applied to the end of the piston of the shock absorber shown in the figure
below causes the two ends of the shock absorber to move toward each other with a speed of
5 ft/s. Determine the head loss associated with the flow of the oil through the channel.
Neglect gravity and any friction force between the piston and cylinder walls.
Gas
p=0
Oil
Piston
Channel
1-in. diameter
200 lb
Solution 5.104
p2 A2
p1 A1
200 lb
From a force balance on the cylinder
p1A1 − p2 A2 = 200 lb
or with p2 = 0,
p1 = 200
lb
lb
lb
= 200
= 3.67 ×10 4 2 = 255 psi
2
A1
π  1  
ft
  ft  
 4  12  


From the energy equation,
V2
p
V2
+ z1 + 1 − hL = 2 + z2 + 2 , where
γ
2g
γ
2g
p1
z1 ≈ z2 ,V2 ≈ 0,V1 = 5
lb
ft
, p1 = 255 psi, and p2 = 0. Assume γ = 50 2 .
s
ft
Thus
2
hL =
p1 V12
+
=
γ 2g
lb
 ft 
5 
ft 2 +  s 
= 734 ft + 0.388ft = 734 ft
ft 
 lb 

2  32.2 2 
 50 3 
s 
 ft 

3.67 ×104
Problem 5.106
Oil (SG = 0.88) flows in an inclined pipe at a rate of 5 ft3 /s as shown in the figure below.
If the differential reading in the mercury manometer is 3ft , calculate the power that the
pump supplies to the oil if head losses are negligible.
6 in.
h
P
H
Oil
12 in.
3 ft
Solution 5.106
(2)
6 in.
h
P
(1)
H
Oil
12 in.
3 ft
Using the control volume shown and the energy equation
2
pout Vout
p
V2
+
+ zout = in + in + zin + hs − hL , we get:
γ
2g
γ
2g
V22
p1 V12
+
+z =
+
+z +h −h
γ oil 2 g 2 γ oil 2 g 1 s
p2
(1)
The power supplied by the pump to the oil is, from the equation
wshaft Wshaft Wshaft
hs =
net in
g
=
net in
mg
=
net in
γQ
:

Wshaft = γ oilQhs = SGoil γ H2OQhs
(2)
net in
Since V =
Q
Q
=
, we get
A πd2
4
ft3
ft
ft
s
= 6.37 and V2 =
=
V1 =
25.5
2
π
s
s
π 1 
(1ft)2
ft


4
42 
5
ft3
s
5
From the manometer equation, we get:
p1 + γ oil H + 3γ Hg − (3 + H + h )γ oil = p2
thus
p1
γ oil
+3
γ Hg
p
−3− h = 2
γ oil
γ oil
(3)
Combining Eqs. (1) and (3), we get,
V22
p1 V12
+ (3 ft )
− (3 ft ) +
+ z2 =
+
+z +h − h
SGoil
γ oil
γ oil 2 g 1 s
2g
p1
SGHg
0
or
2
ft  
ft 

25.5  −  6.37 

 SGHg  V22 −V12
s 
s
 13.6  
hs = z2 − z1 + 3 ft 
− 1 +
= (3ft) 
− 1 +
ft 
2g

 0.88 
 SGoil

2  32.2 2 
s 

hs = 52.9ft
Finally from Eq. (2)
lb   ft3 
ft ⋅ lb

52.9 ft ) = 14500
Wshaft = (0.88)  62.4 3   5
(

s
ft   s 

net in
2
Problem 5.107
The pumper truck shown in the figure below is to deliver 1.5 ft3 /s to a maximum elevation
of 60ft above the hydrant. The pressure at the 4-in. -diameter outlet of the hydrant is
10 psi . If head losses are negligibly small, determine the power that the pump must add to
the water.
60 ft
10 psi
4-in.
diameter
Hydrant
Solution 5.107
To solve this problem, we first use the energy equation
2
pout Vout
pin Vin2
+
+ zout =
+
+ zin + hs − hL for flow from the hydrant exit (1) to the
γ
2g
γ
2g
maximum desired elevation of 60 ft (2) to get hs or in this case, the pump head. With the
pump head, we can get the pump power from the equation hs =
0
0
0
p1 V12
V22
+
+ z2 = +
+ z1 + hs − hL
ρ
ρ 2g
2g
p2
hs = z2 − z1 −
p1 V12
−
ρ 2g
 ft3 
1.5
 ( 4)
s 
ft
Q
Q

17.2
V1 =
=
=
=
2
s
A1  π d12 






4 in.
 4  π
in. 
 12

 ft 
wshaft
Wshaft
net in
net in
g
=
mg
Wshaft
=
net in
γQ
2
lb  
in.2  

ft 
10
144


17.2 


ft 2  
 in.2  
s
−
hs = 60 ft −
lb 
ft 


2  32.3 2 
 62.4 3 
ft 
s 


hs = 32.3ft
lb   ft3  ( 32.3ft )

Wshaft = γ Qhs =  62.4 3  1.5

s  
ft ⋅ lb 
ft  

net in
 550 s ⋅ hp 


Wshaft = 5.48 hp
net in
Problem 5.108
The hydroelectric turbine shown in the figure below passes 8 million gal/min across a head
of 600 ft. What is the maximum amount of power output possible? Why will the actual
amount be less?
600 ft
Turbine
Solution 5.108
(1)
600 ft
(2)
Turbine
From the energy equation,
p1
γ
+ z1 +
V12
p
V2
+ hs − hL = 2 + z2 + 2
2g
γ
2g
where p1 = 0, p2 = 0, and V1 = 0.
Thus,
hs = ( z2 − z1 ) + hL +
V22
2g
And, the power is given by

V22 
=
=
−
+
+
γ
γ
Qh
Q
z
z
h
Wturb
( 2 1 ) L

2 g 

The maximum power would occur if there were no losses ( hL = 0 ) and negligible kinetic
energy at the exit (V2 ≈ 0; large diameter outlet ) .
Thus,
Wturb
= γ Q ( z2 − z1 ) = 62.4
3 
lb 
6 gal   1min   1ft
×
8
10

 ( −600 ft )



min   60 s   7.48gal 
ft3 




⋅
ft
lb
1hp
6
− 6.67 ×108

 = −1.21×10 hp
⋅
ft
lb
s  550

s 

The minus sign is associated with power out.
The actual power will be less due to internal frictional losses and the exit kinetic energy of
the flow.
Problem 5.109
A pump is to move water from a lake into a large, pressurized tank as shown in the figure
below at a rate of 1000 gal in 10 min or less. Will a pump that adds 3 hp to the water work
for this purpose? Support your answer with appropriate calculations. Repeat the problem if
the tank were pressurized to 3, rather than 2, atmospheres.
p = 2 atm
Air
20 ft
Pump
Solution 5.109
p1
γ
+ z1 +
V12
p
V2
+ hs − hL = 2 + z2 + 2 , where p1 = 0, z1 = 0, V1 = 0, and z2 = 20 ft .
2g
2g
γ
Thus,
hs = hL +
(1)
p2
γ
+ z2
Also,
 (1000 gal )   1ft3   1min 
ft3
=
Q=
0.223





s
 (10 min )   7.48 gal   60 s 
so that
lb 

ft ⋅ 

s
(3 hp)  550

hp 
•

Ws

 = 119 ft
hs =
=
γQ 
lb  
ft3 
 62.4 3   0.223 s 
ft  


lb  
in.2 
lb

(a) If p2 = 2 atm = 2  14.7 2   144 2  = 4230 2 , then from Eq. (1)


in.  
ft 
ft

lb
ft 2 + 20 ft = h + 87.8ft
hs = hL +
L
lb 

62.4


ft3 

4230
Thus, if
hL ≤ hs − 87.8ft = 119 ft − 87.8ft = 31.2 ft,the given pump will work for p2 =2 atm.
(b) If p2 = 3atm = 6350
lb
ft 2
, then
lb
ft 2 + 20 ft = h + 87.8ft
hs = hL +
L
lb 

62.4


ft3 

6350
Thus, if this pump is to work
119 ft = hL + 122 ft, or hL ≤ −3 ft
Since it is not possible to have hL < 0, the pump will not work for p2 = 3atm.
Problem 5.110
Water is pumped from the tank shown in the figure (a). The head loss is known to be
V2
1.2
, where V is the average velocity in the pipe. According to the pump manufacturer,
2g
the relationship between the pump head and the flowrate is as shown in the figure (b):
hp = 20 − 2000 Q 2, where hp is in meters and Q is in m3 /s . Determine the flowrate, Q .
hp (m)
20
6m
10
hp = 20–2000 Q2
0
0
0.05
0.10
Q (m3/s)
0.07 m
Pump
(b)
(a )
Solution 5.110
(1)
hp (m)
20
6m
10
hp = 20–2000 Q2
(2)
0
0
0.05
0.10
Q (m3/s)
0.07 m
Pump
(a)
(b)
We want to know the flowrate Q . For the control volume shown, application of the energy
p
V2
p
V2
equation out + out + zout = in + in + zin + hs − hL yields.
γ
2g
γ
2g
2
p2 V22
p1 V1
+
+ z2 = +
γ
γ
2g
2g
0
+ z1 + hs − h
(1)
However
h = 1.2
V2
2g
(2)
and
hs = hp = 20 − 2000 Q 2
(3)
Since Q = V2 A2 , we have from Eq. (2)
h =
1.2  Q 
 
2g  A 
2
(4)
and combining Eqs. (1), (3) and (4), we get:
2
1 Q
2 1.2  Q 
  + z2 = z1 + 20 − 2000 Q −
 
2 g  A2 
2 g  A2 
2
(5)
or
 1

1.2
Q2 
2000
+
+
 = z1 − z2 + 20
 2 gA2 2 gA2
2
2


so
1

2






z1 − z2 + 20

Q=
1
1.2

+
+ 2000 
2
2

2
 πd2 
 2 g  π d2 

2g  2 


 4 
  4 




 

1

2






6 m + 20 m


=

1
1.2
+
+ 2000 

2
2
2
2


m   π ( 0.07 m ) 
m   π ( 0.07 m ) 




 ( 2 )  9.81 2  

( 2 )  9.81 2  
4
4
s  
s  






m3
Q = 0.052
s
Problem 5.111
Water is pumped steadily through the apparatus as shown in the figure below. The pipe
area and gage pressure are shown for both outlet sections (1) and (2). Assume that the 40 °F
water is frictionless and incompressible. Compute the horsepower input to the pump. The
total volume flowrate QT = 1.0 ft3 /s. Da = 4.0 in.
A1 = 0.04 ft2
p1 = 10 psig
patm = 14.7 psi
Q1
1
Q2
25 ft
2
15 ft
A2 = 0.03 ft2
p2 = 15 psig
a
Pump
7 ft
0
Solution 5.111
Denote the pump discharge as point a. Write the mechanical energy equation for a
streamline from point a to point 1. Note ws = 0 and loss = 0 .
Va2
p1 V12
+
+ gza = +
+ gz1
ρ
2
ρ
2
pa
(1)
Next write the mechanical energy equation for a streamline from point a to point 2. Note
ws = 0 and loss = 0
pa
ρ
+
Va2
p V2
+ gza = 2 + 2 + gz2
2
ρ
2
(2)
Equating Eqs. (1) and (2) give
p1 V12
p V2
+
+ gz1 = 2 + 2 + gz2
ρ
2
ρ
2
Since Q1 = AV
1 1 and Q2 = A2V2 , we get
p1
ρ
+
Q12
p2 Q22
gz
+
=
+
+ gz2
1
ρ ∂A22
∂A12
Combining this equation with
QT = Q1 + Q2 or Q1 = QT − Q2
gives
p1
ρ
+
(QT − Q2 )2 + gz = p2 + Q22 + gz , p1 + QT2 − 2QT Q2 + Q22 + gz = p2 + Q22 + gz ,
1
∂A12
ρ
2
∂A22
ρ
1
∂A12
p1 − p2
QT2
1 1
1  2 QT
−
+ g ( z1 − z2 ) + 2

 Q2 + 2 Q2 =
ρ
2  A22 A12 
∂A1
A1
ρ
(3)
The numerical values give (for 40 °F water)


1
1
1
 243
−

= 4
2 0.03ft 2 2
2 2
0.04 ft

 ft


(
) (
)
 ft3 
1.0

s 
QT
625

=
=
2
2
ft ⋅ s
A1
0.04 ft 2
(
p1 − p2
ρ
)
+ g ( z1 − z2 ) +
QT2
∂A12
lb 
in.2 
144


in.2 
ft 2   slug ⋅ ft 

2 
slugs 

lb
s
⋅


1.940 3 
ft 

(10 − 15 )
=
2
 ft3 
1.0

s 
ft 2
 32.2 ft 

=
263
+  2  ( 25 − 15 ) ft+
2
s2
 s

2 0.04 ft 2
(
Equation (3) is then
243
ft
Q2 +
4 2
625
ft 2
Q2 = 263 2
ft ⋅ s
s
or
ft
Q22 + 2.57
3
s
Q2 − 1.08
ft 6
s2
=0
Solving for Q2 gives
Q2 =
−2.57 ± 2.572 + 4 (1.08 ) ft 3
2
s
= 0.358
ft3
ft3
, − 2.94
.
s
s
The meaningful values for this problem are
Q2 = 0.358
ft3
s
and
ft3
Q1 = (1.0 − 0.358)
s
)
∂A22
2
Q1 = 0.642
ft3
s
The pump input power is found by writing the mechanical energy equation for the pump on
a streamline from point 0 to point a,
V02
pa Va2
+
+ gz0 + ws =
+
+ gza + gh L
2
2
ρ
ρ
p0
The problem statement gives V0 ≈ 0 and loss = 0. Then
 pa − p0 Va2

ws = + 
+
+ g ( za − z0 )  . (3)
2
 ρ

The velocity Va is computed from
 ft3 
4 1.0


s 
QT 4QT
ft

Va =
=
=
= 11.5 .
2
2
Aa π Da
s
 4 
π  ft 
 12 
The pressure pa is found using the mechanical energy equation (2).
V 2 −Va2

pa = p2 + ρ  2
+ g ( z2 − za ) 
2


where

ft3 
 0.358

s 
Q2 
ft
V2 −
=
= 11.9
2
A2
s
0.03ft
For 40 °F water, ρ = 1.940
slugs
ft3
.so
.
 ft 2
slugs   (11.92 − 11.52 )

pa = 15 psig + 1.940 3  
+ 32.2(15 − 7)  2
2
ft  

 s
 lb ⋅ s 2   ft 2 
*
= 18.5 psig
 slug ⋅ ft   144 in.2 



Then Eq. (3) gives
2
 

lb  
in.2  
 18.5 2  144 2   11.5 ft 

in.  
ft  
ft 2   lb ⋅ s 2 
ft ⋅ lb
s
 
ws = + 
+
+ 32.2(7) 2  
= +1670

2 
2
slug

s   slug ⋅ ft 
  1.940 slugs   lb ⋅ s 
3
 

ft   slug ⋅ ft 
The mass flowrate is
•
slugs   ft3 
slugs

m = ρQT = 1.940 3  1.0
= 1.940

s 
s
ft  

The pump input power is
•
•
slugs  
ft ⋅ lb   s ⋅ hp 

Ws = m ws = 1.940

  +1670

s 
slug   550 ft ⋅ lb 

•
Ws = +5.89 hp
Problem 5.112
Water is pumped from the large tank as shown in the figure below. The head loss is known
to be equal to 4V 2 / 2 g and the pump head is hp = 20 − 4Q 2, where hp is in ft when Q is in
ft 3 /s. Determine the flowrate.
13 ft
V
Q
Pump
Pipe area = 0.10 ft2
Solution 5.112
V2
p
V2
+ z1 + 1 + hs − hL = 2 + z2 + 2 , where p1 = p2 = 0, z1 = 13ft, z2 = 0, hs = hp , and V1 = 0.
γ
γ
2g
2g
p1
Thus,
z1 + hp − hL =
V22
2g
(1)
Also,
Q
 
2
2
A
V
V2
hL = 4
=4
=4 2 
2g
2g
2g
2
since V2 =
Q
A2
Hence, Eq. (1) becomes
2
Q 
Q 
 
 
A2 
A

2
z1 + (20 − 4Q ) − 4
= 2
2g
2g
2
or
 5   2
ft
ft3
2



+
Q
=
+
z
g
A
4
20
,
where
,
ft
,
and
Q



1
2
2
s
s2
 2 gA2  
Thus, with the given data

 

 
5

 + 4  Q 2 = 20 + 13ft
ft




2 2
 2  32.2 s 2  (0.1ft )  
 

or
Q = 1.67
ft3
s
Problem 5.113
Water flows by gravity from one lake to another as sketched in the figure below at the
steady rate of 80 gpm. What is the loss in available energy associated with this flow? If this
same amount of loss is associated with pumping the fluid from the lower lake to the higher
one at the same flowrate, estimate the amount of pumping power required.
50 ft
Solution 5.113
(a)
50 ft
(b)
gal
ft3
min
Q=
= 0.178
s 
gal 
s

 60
 7.48 3 
ft 
 min 
80
For the flow from sections (a) to (b), the energy equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss leads to
2
2
ρ
ρ
net in




ft 
lb
ft ⋅ lb

 = 1610
loss = g ( za − zb ) =  32.2 2  ( 50 ft ) 1
2
slug

s 
ft 

 slug ⋅

s 

For pumped flow from sections (b) to (a), the energy equation (see above) yields






ft 
lb



32.2
50
ft
1
(
)


slugs  
ft3  


s2 
ft 2  
Wshaft = ρQ [ g ( za − zb ) + loss ] = 1.94 3   0.118  
 slug ⋅

s 
s  

ft  

net in

ft ⋅ lb


 +1610 slug



or
Wshaft = 1110
net in
ft ⋅ lb
= 2.02 hp
s
Problem 5.114
The turbine shown in the figure below develops 100 hp when the flowrate of water is
20 ft3 /s . If all losses are negligible, determine (a) the elevation h , (b) the pressure difference
across the turbine, and (c) the flowrate expected if the turbine were removed.
p3
p4
h
12 in.
T
12 in.
Free jet
Solution 5.114
(1)
CVA
p3
CVB
p4
T
(4)
12 in.
h
(2)
(3)
12 in.
Free jet
(a) Using control volume A and the energy equation
2
pout Vout
pin Vin2
+
+ zout =
+
+ zin + hs − hL
γ
2g
γ
2g
we get:
2
p2 V22
p1 V1
+
+ z2 = +
γ
2g
γ
0
2g
+ z1 + hs − h
0
(1)
For a turbine, hT = − hs and from the equation hs =
ft ⋅ lb 
s ⋅ hp 
net out

=
= 44.1ft
γQ
lb   ft3 

 62.4 3   20 s 
ft  


Wshaft
hT =

(100 hp )  550
wshaft
Wshaft
net in
net in
g
=
mg
Wshaft
=
net in
γQ
, we get:
Since Q = AV , we have
ft3
ft3
20
ft
Q
s =
s
=
=
25.5
V2 =
2
s
A2
π d22




π 12 in.
4


4  12 in. 
 ft 
20
Then from Eq. (1)
2
ft 

25.5 

2
V
s
z1 − z2 = h = 2 − hs = 
+ 44.1ft = 54.1ft
ft 
2g

2  32.2 2 
s 

(b) For control volume B, the energy equation yields
lb 
lb

p3 − p4 = γ hT =  62.4 3  ( 44.1ft ) = 2.75 2
ft 
ft

(c) Since Q = VA = V2 A2, if we knew value of V2 with the turbine removed, we could
calculate Q with the turbine removed. Without the turbine, Eq. (1) reduces to
V22
= z1 − z2 = h
2g
and
ft 
ft

V2 = 2 gh = 2  32.2 2  ( 54.1ft ) = 59
s
s 

Thus
2


π d22
ft 2
π  12 in.   ft 
=
59
46.3
Qw / o =
V2 = 
 

4
4  12 in.   s 
s
turbine
 ft 
Problem 5.115
The figure below shows a pump testing setup. Water is drawn from a sump and pumped
through a pipe containing a valve. The water is discharged into a catch tank sitting on a
scale. During a test run, 800 lb of water is collected in the catch tank in 15 s. The pump
power input to the fluid during this period is 700 ft ⋅ lb/s. Calculate the water velocity in the
pipe and the mechanical energy loss (ft ⋅ lb/lbm) in the pipe and valve.
4-in. I.D. pipe
Valve
5 ft
Catch
tank
60°F H2O
Sump
Solution 5.115
For 60 °F water, ρ = 62.4 lbm ft3
so
 800 lbm 


15s 
Q m

V= =
2
A ρA 
lbm  π  4 
 62.4 3  4  12 ft 
ft  


or
•
V = 9.79
ft
s
We next write the mechanical energy equation from the water level (1) in the sump to the
4-in. pipe discharge (2). For steady state conditions,
p1 V12
p V2
+
+ gz1 + w p = 2 + 2 + gz2 + loss.
ρ
ρ
2
2
Now p1 = p2 = patm and V1 ≈ 0. Also z2 − z1 = 5.0 ft and
lb 

700 ft ⋅ 

ft ⋅ lb
W
s 
= 13.1
wp = • = 
lbm
 800 lbm 
m 

 15 s 
•
Then
V22
loss = w p −
− g ( z2 − z1 )
2
2


ft 
  9.79 

ft ⋅ lb  
lb ⋅ s2
s  32.2 ft
= 13.1
−
+ 2 (5 ft) 
 32.2 ft ⋅ lbm
lbm 
2
s




loss = 6.61
ft ⋅ lb
lbm
Problem 5.116
Water is to be moved from one large reservoir to another at a higher elevation as indicated
in the figure below. The loss of available energy associated with 2.5 ft 3 /s being pumped
from sections (1) to (2) is loss = 61V 2 / 2 ft 2 /s2 , where V is the average velocity of water in
the 8-in. inside-diameter piping involved. Determine the amount of shaft power required.
Section (2)
8-in. insidediameter pipe
50 ft
Section (1)
Pump
Solution 5.116
For the flow sections (1) to (2), the equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss
2
2
ρ
ρ
net in

V2 
Wshaft = ρQ  g ( z2 − z1 ) + loss  = ρQ  g ( z2 − z1 ) + 61 
2 

net in

leads to
(1)
From the volume flowrate, we obtain

ft3 
 2.5

s 
Q
Q
ft

V= =
=
= 7.162
2
2
A πD
s


π  8 in. 
4


4  12 in. 
 ft 
Thus, from Eq. (1)
2

ft   



(61)
7.162



3 




slugs 
ft 
ft 
1lb
1

s 



+
32.2
(50
ft)
Wshaft =  1.94 3   2.5





ft  
ft ⋅ lb 
s  
2
ft  
s2 
net in 
⋅
slug
550


2 
s ⋅ hp 
s 

 
or
Wshaft = 28 hp
net in
Problem 5.117
Determine the volume flowrate and minimum power input to the water pump in the figure
below. Determine the actual power if the hydraulic efficiency is 75% and losses in the
motor and bearings are negligible.
patm = 14.7 psia
p2
p2 = 49.5 psia
d2 = 4.0 in.
d1 = 6.0 in.
p1
p1 =
13.0 psia Q
16.0 ft
8.0 in.
Solution 5.117
The minimum pump input power occurs for frictionless flow. Assume constant fluid
density. Apply Bernoulli’s equation from the free water surface (0) to the pump inlet pipe at
location 1, 8 in. above water surface.
p0 V02
p V2
+
+ gz0 = 1 + 1 + gz1
ρ
2
ρ 2
Now V0 = 0 so
p −p

V1 = 2  0 1 + g ( z0 − z1 )  .
 ρ

Assuming 60 °F water


lb 
in.2 
(14.7
13.0)
144
−




in.2 
ft 2  
ft   8  
ft

V1 = 2 
+  32.2 2   − ft   = 14.5
2 
s

s   12  

 1.938 slugs   lb ⋅ s 
 

ft3   ft ⋅ slug 
and
2
ft 
6  
π  ft  14.5 
2
π d1 V1
ft3
s
 12  
2.85
=
=
=
=
Q = AV
Q
1 1
4
4
s
The pump input power is found by writing the mechanical energy equation from the free
water surface (0) to point 2 at the pump discharge.
V02
p2 V22
+
+ gz0 + ws =
+
+ gz2 + loss
ρ
ρ
2
2
p0
Now V0 = 0 and the minimum pump input power occurs if loss = 0 . This gives
 p − p0 V22

ws =  2
+
+ g ( z2 − z0 ) .
2
 ρ

The continuity equation gives
2
2
A 
d  
ft   6.0 in. 
ft
V2 = V1  1  = V1  1  = 14.5  
 = 32.6 .
s   4.0 in. 
s
 A2 
 d2  
Then
2


lb 
in.2 
 (49.5 − 14.7) 2 144 2   32.2 ft 

2
 lb ⋅ s 2  
in. 
ft  
ft 
s   lb ⋅ s  

ws = 
+

 +  32.2 2  (16.0 ft ) 

slugs 
2
ft ⋅ slug  
ft ⋅ slug  

s





1.938 3 
ft 



ft ⋅ lb
= 3630
slug
Then
slugs  
ft3  
ft ⋅ lb 

1.938
2.85
3630





3
•
•
s  
slug 
ft  

Ws = mws = ρQws =

ft ⋅ lb 
 550 s ⋅ hp 



Ws = 36.5 hp
Problem 5.118
A pump moves water horizontally at a rate of 0.02 m3 /s. Upstream of the pump where the
pipe diameter is 90 mm, the pressure is 120 kPa. Downstream of the pump where the pipe
diameter is 30 mm, the pressure is 400 kPa. If the loss in energy across the pump due to
fluid friction effects is 170 N ⋅ m/kg, determine the hydraulic efficiency of the pump.
Solution 5.118
The efficiency of the pump,η , is
ideal work required actual work required − loss
η=
=
=
actual work required
actual work required
wshaft − loss
net in
wshaft
net in
To determine wshaft , we use the equation
net in
2
Vout
pin Vin2
+
+ gzout =
+
+ gzin + wshaft − loss
ρ
ρ
2
2
net in
pout
to obtain
wshaft =
pout − pin
net in
ρ
2
Vout
−Vin2
+
+ loss
2
(1)
From the volume flowrate, we obtain

m3 
0.02


s 
m
Q
Q

Vout =
=
=
= 28.29
2
2
s
Aout π Dout
( 0.030 m )
π
4
4
Also, from mass conservation
Vin = Vout
2
Dout
m  (0.030)2
m

=
= 3.143
28.29


2
2
s  (0.090)
s
Din 
Thus from Eq. (1), we obtain
2
2

N
N   28.29 m  −  3.143 m   


 

 400000 2 − 120000 2  
s
s






N
N⋅m
 1
m
m +
wshaft = 

 + 170
kg 
2
kg

net in
 kg ⋅ m 
 999 3 
2
s 


m 
or
N⋅m
wshaft = 846
kg
net in
Then
846
η=
N⋅m
N⋅m
− 170
kg
kg
= 0.799
N⋅m
846
kg
Problem 5.119
Water is to be pumped from the large tank shown in the figure below with an exit velocity
of 6 m/s. It was determined that the original pump (pump 1) that supplies 1kW of power to
the water did not produce the desired velocity. Hence, it is proposed that an additional
pump (pump 2) be installed as indicated to increase the flowrate to the desired value. How
much power must pump 2 add to the water? The head loss for this flow is hL = 250Q 2 ,
where hL is in m when Q is in m3 /s .
V = 6 m/s
Nozzle area = 0.01 m2
Pipe area = 0.02 m2
Pump
#2
Pump
#1
2m
Solution 5.119
V = 6 m/s
Nozzle area = 0.01 m2
Pipe area = 0.02 m2
(2)
Pump
#2
Pump
#1
2m
(1)
V12
p2
V22
+ z1 +
+ hS − hL =
+ z2 +
, where p1 = p2 = 0, z1 = 0, z2 = 2 m, and V1 = 0 .
2g
2g
γ
γ
Thus,
p1
m
m
V22
 m
hS = z2 + hL +
, where V2 = 6 so that Q = A2V2 = 0.01m2  6  = 0.06
s
s
2g
 s 
Note: hS = hpump1 + hpump2
Thus, with hL = 250Q2 = 250( 0.06)2 = 0.90 m, it follows that
2
 m
6 
s 
hS = 0.90 m + 2 m + 
= 4.73 m
m

2  9.81 2 
s 

so that
m3 
N⋅m

3 N 
WS = γ QhS =  9.80 ×10
= 2.78kW
0.06
 ( 4.73 m ) = 2.78 × 10
3 

s 
s

m 
•
Therefore,
WS = Wpump1 + Wpump2 = 2.78kW, with Wpump1 = 1kW
Hence,
Wpump2 = 2.78kW − 1kW = 1.78kW
Problem 5.120
Curtain of air An air curtain is produced by blowing air through a long rectangular nozzle
to produce a high-velocity sheet of air, or a “curtain of air.” This air curtain is typically
directed over a doorway or opening as a replacement for a conventional door. The air
curtain can be used for such things as keeping warm air from infiltrating dedicated cold
spaces, preventing dust and other contaminants from entering a clean environment, and
even just keeping insects out of the workplace, still allowing people to enter or exit. A
disadvantage over conventional doors is the added power requirements to operate the air
curtain, although the advantages can outweigh the disadvantage for various industrial
applications. New applications for current air curtain designs continue to be developed. For
example, the use of air curtains as a means of road tunnel fire security is currently being
investigated. In such an application, the air curtain would act to isolate a portion of the
tunnel where fire has broken out and not allow smoke and fumes to infiltrate the entire
tunnel system. (See Problem 5.120.)
The fan shown in the figure below produces an air curtain to separate a loading dock from
a cold storage room. The air curtain is a jet of air 10 ft wide, 0.5ft thick moving with speed
V = 30ft/s . The loss associated with this flow is loss = K LV 2 / 2, where K L = 5. How much
power must the fan supply to the air to produce this flow?
Fan
V = 30 ft/s
Air curtain
(0.5-ft thickness)
Open door
10 ft
Solution 5.120
(1)
Fan
(2)
V = 30 ft/s
Air curtain
(0.5-ft thickness)
Open door
10 ft
p1
γ
+ z1 +
V12
p
V2
+ hs − hL = 2 + z2 + 2 ,
2g
γ
2g
where p1 ≈ p2 ≈ 0, z1 ≈ z2 , V1 ≈ 0 and hL =
Thus,
V2
loss
=5 2 .
g
2g
2
 ft 
3  30 
2
2
2
2
V
V
V
V
s
= 83.9 ft
hs = hL + 2 = 5 2 + 2 = 3 2 = 
ft 
2g
2g 2g
g

 32.2 2 
s 

Hence,
•
slugs  
ft 

 ft 
WS = γ QhS = ρ gA2V2 hS =  0.00238 3   32.2 2  (10 ft)(0.5 ft)  30  (83.9 ft)
 s
ft  
s 



ft ⋅ lb  1hp 
= 964


s  550 ft ⋅ lb 
s 

= 1.75 hp
Problem 5.121
When the pump shown in the figure below is stopped, there is no flow through the system
and the spring force is zero. With the pump running, a 6-in.-diameter jet leaves the pipe,
and the spring force is 420 lb . The water surface elevation in the tank is constant.
Determine the water flowrate and the power consumed by the pump. Assume quasi-steady
flow.
6-in. diameter
Pump
Jet
3.0'
Spring
30°
1.0'
Frictionless rollers
Solution 5.121
Write the linear momentum equation in the horizontal or x-direction for a control volume
enclosing the tank. Positive is to the right.
•
•
 Fx = mVx,out − mVx,in
where
 Fx = Fs = spring force acting to right on control volume,
•
•
mVx,out = ρ AV 2 cos30 , and mVx,in = 0
where V is the jet velocity leaving the discharge pipe and A is discharge pipe flow area.
Assuming 60 °F water,
 ft ⋅ slug 
(420 lb) 
2 
Fs
lb
⋅
s


=
V=
2
ρ A cos30
slugs   π   6 

1.938 3   4   12 ft  cos30
ft    


ft
= 35.7
s
The water flowrate is
2
ft   π   6 
ft3

Q = VA =  35.7     ft  = Q = 7.01
s   4   12 
s

The pump power is found by first writing the mechanical energy equation from the water
surface (0) to the jet discharge (d).
For frictionless flow
V02
pd Vd2
+
+ gz0 − wS =
+
+ gzd
ρ
2
ρ
2
p0
Now p0 = pd = patm and V0 ≈ 0 to get
2
ft 

35.7 
2

V
ft 
ft 2
s 
−wS = d + g ( zd − z0 ) = 
+  32.2 2  ( −2 ft) = 573 2
2
2
s 
s

The pump power is
slugs  
ft3 
ft 2  lb ⋅ s2 

−
1.938
7.01
573





•
•
s 
ft ⋅ slug 
ft3  
s2 



WS = mwS = ρQwS =

ft ⋅ lb 
 550 hp ⋅ s 


•
WS = −14.4 hp
Problem 5.122
Air flows past an object in a pipe of 2-m diameter and exits as a free jet as shown in the
figure below. The velocity and pressure upstream are uniform at 10 m/s and 50 N/m 2 ,
respectively. At the pipe exit, the velocity is nonuniform as indicated. The shear stress along
the pipe wall is negligible. (a) Determine the head loss associated with a particle as it flows
from the uniform velocity upstream of the object to a location in the wake at the exit plane
of the pipe. (b) Determine the force that the air exerts on the object.
Air
4 m/s
Wake 1-m dia.
2-m-dia.
12 m/s
Exit
p = 50 N/m2 V = 10 m/s
Solution 5.122
(1)
(2)
Air
2-m-dia.
Rx
Wake 1-m dia.
4 m/s
12 m/s
Exit
p = 50 N/m2 V = 10 m/s
(a) To determine the loss suffered by a fluid particle as it flows from (1) to a location in
the wake at (2), we apply the energy equation
2
pout Vout
p
V2
+
+ zout = in + in + zin + hs − hL
γ
2g
γ
2g
to that particle flow to get:
0
p2
γ
0
Wshaft net in
V2
p V2
+ 2 + z2 = 1 + 1 + z1 +
− hL
g
2g
γ 2g
(1)
or
hL =
p1 V12 V22
+
−
γ 2g 2g
and
2
2
N   m

 m
 50 2  10 
4 
s 
s 



m
hL =
+
− 
= 8.45 m
m
m
 N 


12 3  2  9.81 2  2  9.81 2 
 m 
s 
s 


To determine the head loss associated with the entire flow across the object we use the
wshaft
2
2
pout αoutV out
pin α inV in
net in
+
+ zout =
+
+ zin +
− hL
nonuniform flow energy equation
2g
2g
g
γ
γ
for flow from (1) to (2) through the control volume shown in the sketch to get:
0
0
p2
+
γ
2
α 2V 2
2g
+ z2 =
p1
γ
+
2
α1V 1
2g
+ z1 +
Wshaft net in
g
− hL (2)
V2
A 2 ρ V ⋅ ndA
α
=
, we get:
From
• V2
m
2
αV
2
2g
=
V2
 2 g ρ V ⋅ ndA
•
m
Eq. (2) becomes
hL =
V2
 2 g ρ V ⋅ ndA
=
ρVA
V2
 2 g ρ V ⋅ ndA
A
p1 V12
2
+
−
γ 2 g ( ρVA) m + ( ρVA)
4
s
12
m
s
or
V 3 m A m +V 3m A m 
3
3 
p
1  12 s 12 s
s
s
−
hL = 1 +


γ 2 g 2 g V m A m +V m A m 
12
12 
 4 s 4 s
s
s 
V12
and
2
  m 3  (2 m)2 − (1m)2   m 3 π (1m)2 
N   m

 12  π 

 +4 
 50 2   10 
s
4
4


1

  s 


s 
m




hL =
+
−


2
2
2
N
m
m





   m  π (1m)  m   (2 m) − (1m)  
12
2
9.81
2
9.81


4
+  12  π 




 
 m3 
s2 
s2    s 


4
4
 s  
 

hL = 2.58 m
(b) To determine the force that the air puts on the object, Rx , we use the horizontal
component of the linear momentum equation to get:
ρV12 A1 + ρV 2 m A
12
s
2
m + ρV m A m = p1A1 − Rx
12
4
4
s
s
s
and thus


Rx = p1A1 + ρV12 A1 − ρ V 2 m A m +V 2m A m 
 12 12
4
4 
s
s
s
s 

so
2
N  π (2 m)2 
kg  m  π (2 m)2  N ⋅ s2 

Rx =  50 2 
+ 1.23 3 10 
1

4
4
 m 

m  s 
 m ⋅ kg 
2
2


2 
2
kg  m  π (2 m) − (1m)   m  π (1m)2   N ⋅ s2 
− 1.23 3 12 
+4 

 1
4
4  m ⋅ kg 
 s 
m  s 


and
Rx = 110 N
Problem 5.123
Near the downstream end of a river spillway, a hydraulic jump often forms, as illustrated in
the figure below. The velocity of the channel flow is reduced abruptly across the jump.
Using the conservation of mass and linear momentum principles, derive the following
2
2
h
 h  2V h
expression for h2 , h2 = − 1 +  1  + 1 1
g
2
2
The loss of available energy across the jump can also be determined if energy conservation
g ( h2 − h1 )3
is considered. Derive the loss expression jump loss =
4 h1h2
h1
h2
V1
Solution 5.123
Section (2)
B
Section (1) h
1
h2
A
V1
Application of the horizontal component of the linear momentum equation
∂
Vρ d V +  Vρ V ⋅ ndA =  Fcontents of the to the water in the control volume from
CS
∂t CV
control volume
sections (1) to (2) leads to, for unit width of flow,
−Rx +
γ h12
2
−
γ h22
2
= −V1ρ hV
1 1 + V2 ρ h2V2
(1)
Since the jump occurs over a short distance we assume Rx = 0. Also from conservation of
mass Q = AV
1 1 = A2V 2 , we obtain
h
V2 = V1 1
h2
(2)
Combining Eqs. (1) and (2), we obtain



 h2 
1
1−   =
− 1


h
gh


h
1
 1
 2  
  h1  
2

2V12 
Or


 h2   h2  2V12 
=
1−
1+



h1  

h1 
gh1
h 
1− 2 
h1 
 h2 
 
 h1 
and
2
 h2   h2  2V12
=0
  + −
h
h
gh
1
1
1
   
(3)
From Eq. (3), we obtain
h2
=
h1
−1 ± 1 +
8V12
gh1
2
or
2
2
h1
 h1  2V1
h2 = − +   +
h1
g
2
2
The other quadratic root is not meaningful.
p
V2
p
V2
Application of the energy equation out + out + gzout = in + in + gzin + wshaft − loss to
2
2
ρ
ρ
net in
the flow from point A to point B shown on the sketch above leads to
Pump loss =
VA2 −VB2
V 2 −V22
+ g ( zA − zB ) = 1
+ g ( h1 − h2 )
2
2
Combining Eqs. (2), (3), and (4), we obtain
2
2
gh1  h2  h2    h1  
  +  1 −    + g ( h1 − h2 )
Pump loss =
h1    h2  
4  h1 



or
Pump loss =
g
( h2 − h1 )3
4h2 h1
(4)
Problem 5.124
Water flows steadily down the inclined pipe as indicated in the figure below. Determine the
following: (a) the difference in pressure p1 − p2 , (b) the loss between sections (1) and (2),
(c) the net axial force exerted by the pipe wall on the flowing water between sections (1)
and (2).
Flo
w
5 ft
Section (1)
6 in.
30°
Section (2)
6 in.
Mercury
Solution 5.124
Flo
w
p1 A
1
Section (1)
W
D = 6 in.
5 ft = ℓ
Control
volume
30°
Rx
30°
Section (2)
p2 A2
6 in.
Mercury
(a) The difference in pressure, p1 − p2 , may be obtained from the manometer with the
fluid statics equation



(6 in.) 
(6 in.)
 + γ Hg
p1 − p2 = −γ H 2O (5 ft)sin30 +
 in.  
 in. 

12  
 12 

 ft  
 ft 
or
lb 
lb 
lb


p1 − p2 = −  62.4 3  (5 ft)sin30 + (0.5 ft)  + (13.6)  62.4 3  (0.5 ft) = 237 3


ft 
ft 
ft


and
lb
1
p1 − p2 = 237 3
= 1.65 psi
ft 
in.2 
144 2 
ft 

(b) The loss per unit mass between sections (1) and (2) may be obtained with the
p
V2
p
V2
equation out + out + gzout = in + in + gzin − loss
Thus,
ρ
ρ
2
2
loss =
p1 − p2
ρ
+
V12 −V22
lb 
1

+ g ( z1 − z2 ) =  237 3 
2
ft  1.94 slugs 



ft 2 




ft 
lb 

+  32.2 2  ( 5 ft ) sin30 1

s 

 slug ⋅ ft 
s2 

(
)
or
loss = 203
ft ⋅ lb
slug
(c) The net axial force exerted by the pipe wall on the flowing water may be obtained by
using the axial component of the linear momentum equation
∂
Vρ d V +  Vρ V ⋅ ndA =  Fcontents of the
CS
∂t CV
control volume
Thus for the control volume shown above
Rx = −
π D2
4
( p1 − p2 ) −
γπ D 2
4
( ) sin30 = −
π D2 
4 
( p1 − p2 ) + γ sin30 

or
2


π  6 in.  
lb
lb

Rx = − 
237 2 + 62.4 3 (5 ft)sin30 


4  12 in.  
ft
ft

 ft 
and
Rx = −77.2 lb = 77.2 lb opposite to flow direction.
Problem 5.125
When fluid flows through an abrupt expansion as indicated in the figure below, the loss in
available energy across the expansion, loss ex , can be expressed as
2
 A1  V12
lossex = 1 −

 A2  2
where A1 = cross-sectional area upstream of expansion, A2 = cross-sectional area
downstream of expansion, and V1 = velocity of flow upstream of expansion. Derive this
relationship.
Section (1)
Section (2)
Solution 5.125
Section (1)
Alternate
location of
section (1)
Section (2)
Applying the energy equation
pout
ρ
+
2
Vout
p
V2
+ gzout = in + in + gzin + wshaft − loss to the flow
2
2
ρ
net in
from sections (1) to (2), we obtain
p1 − p2 V12 −V22
+
lossex =
ρ
2
(1)
Applying the axial direction component of the linear momentum equation
∂
Vρ d V +  Vρ V ⋅ ndA =  Fcontents of the to the fluid contained in the control volume
CS
∂t CV
control volume
from sections (1) to (2), we obtain
Rx + p1A1 − p2 A2 = −V1ρ AV
1 1 + V2 ρ A2V2
(2)
Now, if we consider section (1) as occurring at the end of the smaller diameter pipe (the
beginning of the larger diameter pipe) as indicated in the sketch above, Eq. (1) still yields
the expansion loss and Eq. (2) becomes
Rx + p1A2 − p2 A2 = −V1ρ AV
1 1 + V2 ρ A2V2
(3)
Note that with section (1) positioned at the end of the smaller diameter pipe, p1 acts over
area A2 . Also, because of the jet flow from the smaller diameter pipe into the larger
diameter pipe, the value of Rx will be small enough compared to the other terms in Eq. (3)
that we can assume Rx = 0 . Then from Eq. (3)
p1 − p2
ρ
= V22 −V12
A1
A2
(4)
Combining Eqs. (1) and (4), we obtain
loss ex = V22 −V12
A1 V12 −V22
+
2
A2
From conservation of mass Q = AV
1 1 = A2V2 , we have
V2 = V1
A1
A2
(6)
Combining Eqs. (5) and (6), we get
A 
lossex = V12  1 
 A2 
2
A 
− V12  1  +
 A2 
A 
V12 −V12  1 
 A2 
2
or
2
2
 A1  
V12   A1 
A1
2 
+1− 
lossex =
 −2
 
2   A2 
A2
A2  



and
V2 
A 
lossex = 1 1 − 1 
2  A2 
2
2
Problem 5.126
Water (60 °F) flows through an annular space formed by inserting a 1-in.-radius solid
cylinder into a 1.5-in.-radius tube. The following axial velocities were measured in the
annular space.
(r - ri)/(r0 - ri)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
u (ft/s)
12
23
28
33
34
31
28
21
10
Assume that the no-slip condition (u = 0) exists at the solid boundaries. What are the rates
of mass, momentum, and kinetic energy flow through the annular space?
Solution 5.126
Perform required integrations using trapezoidal rule.
•
Ro
Ro
Ri
Ri
m =  ρ udA =  ρ u 2π rdr = 2πρ  urdr
for each interval:
• rnh
m
 Δr

= 2πρ 
ur n +1 + ur n 
2

rn
(
• rnh
(
= πρΔr ur n +1 + ur n
m
rn
• rnh
M
rn
•
KE
)
rnh
(
= πρΔr u 2 r
n +1
)
+ u2r
n
)
1
1

= πρΔr  u3r
+ u3r 
n
n
+
1
2
2

rn
Ri =
Ro =
dr =
1 in.
1.5 in.
0.004167 ft
Rho=
pi=
62.4
3.14159
For each interval:
mdot = pi*Rho*dr*(ur(n+1) + ur(n))
Mdot = pi*Rho*dr*(u2r(n+1) + u2(n))
KEdot = pi*Rho*dr*(u3r(n+1) + u3r(n))/2
r (ft)
0.0833
0.0875
0.0917
0.0958
0.1000
0.1042
0.1083
0.1125
0.1167
0.1208
0.1250
u (ft/s)
0
12
23
28
33
34
31
28
21
10
0
Totals
mdot
(lbm/s)
Mdot
(ft-lbm/s2)
0.8577
2.5798
3.9139
4.8873
5.5884
5.6360
5.3161
4.5742
2.9882
0.9870
3.733E+01
10.2918
49.9005
100.9786
150.3209
187.3089
183.3950
157.0800
114.0680
51.8949
9.8698
1.015E+03
KEdot
(ft2-lbm/s3)
61.7511
517.2505
1314.6782
2326.8699
3139.7763
2990.1599
2326.6760
1449.8642
490.6122
49.3491
1.467E+04
lbm/ft3
Problem 5.127
Find the acceleration of the cart shown in the figure below as a function of the water height
in the cart, which varies with time. The initial total mass is m0, and the fluid density is ρ0 .
Assume frictionless bearings, a frictionless surface, constant fluid density, uniform velocity
over area AN , all the fluid in the cart has the cart velocity, no air drag, and A  AN .
A
g
AN
h0 = Initial fluid height
ρ0
Solution 5.127
Apply the linear momentum equation in the horizontal or x-direction for a control volume
moving with the cart. The observer and coordinate system are fixed on the ground.
•
•
d
(M x,CV ) + M x,out − M x,in =  Fx
dt
Now,
+


M x,out = ρ0VN AVe and M x,in = 0
where Ve is the absolute exit fluid velocity and VN is the exit fluid velocity relative to the
cart. Because of the frictionless surface and no air drag,
 Fx = 0.
Denoting the mass of the cart and the fluid in the cart at time t by m and since all the fluid
in the cart has the cart velocity Vx , we get
d (M x,CV )
dt
=
d ( mVx )
dt
Ve and Vx are related by Ve = Vx −VN so the linear momentum equation is now
d ( mVx )
+ ρ0VN A(Vx −VN ) = 0
dt
or
m
dVx
dm
+ Vx
+ ρ0 AVNVx − ρ0 AVN2 = 0 .
dt
dt
Applying the integral continuity equation to the control volume for our observer gives

d ( mCV ) 
= min − mout.
dt
Since mCV = m, this equation is
dm
= 0 − ρ 0 ANVN .
dt
The linear momentum equation is then
m
dVx
+Vx ( − ρ0 ANVN ) + ρ0 AVNVx − ρ0 AVN2 = 0,
dt
m
dVx
− ρ0 AVN2 = 0,
dt
or a =
dVx ρ0 AVN2
=
dt
m
This is one possible from of the answer. If we assume the flow inside the cart and through
the nozzle is frictionless and assume we can write Bernoulli’s equation for an observer on
the cart, we get
VN = 2 gh
If we assume A  AN . The mass in the cart at any time t is
m = m0 − ρ0 A( h0 − h )
where h is the fluid height above the exit nozzle at time t . The cart acceleration is then
a=
dVx
2 ρ0 gAh
=
dt
m0 − ρ0 A( h0 − h )
Comment We have a small inconsistency. We assumed previously all the fluid in the cart
had the cart velocity Vx but some of the fluid in the cart must move relative to the cart in
order to leave through area AN . This contradiction is unavoidable without greatly
increasing the complexity of the solution.
Problem 5.128
Two water jets collide and form one homogeneous jet as shown in the figure below. (a)
Determine the speed, V , and direction, θ , of the combined jet. (b) Determine the loss for a
fluid particle flowing from (1) to (3), from (2) to (3). Gravity is negligible.
V
0.12 m
(3)
θ
V2 = 6 m/s
(2)
90°
0.10 m
(1)
V1 = 4 m/s
Solution 5.128
V
Control
volume
0.12 m
(3)
θ
V2 = 6 m/s
(2)
90°
0.10 m
(1)
V1 = 4 m/s
For the water flowing through the control volume sketched above, the x- and y- direction
components of the linear momentum equation are
−V2 ρV2 A2 + (V3 cos θ ) ρV3A3 = 0
(1)
and
−V1ρV1A1 + (V3 sin θ ) ρV3 A3 = 0
(2)
From the conservation of mass principle, we get
− ρV1A1 − ρV2 A2 + ρV3A3 = 0
(3)
Combining Eqs. (1) and (2), we obtain
 m  π ( 0.1m )
V1
2
4 
V1 A1
s 
4
4
tan θ = 2 =
= 
= 0.3086
2
2
π d2  m  π ( 0.12 m )2
V2 A2
V2
6 
4
4
 s 
π d12
so
θ = tan−1 0.3086 = 17.2
2
2
Now, combining Eqs. (1) and (3), we get
−V22 ρ A2 +V3 cos θ ( ρV1A1 + ρV2 A2 ) = 0
or
V3 =
V22 A2
V22 d22
=
cos θ (V1A1 + V2 A2 ) cos θ (V1d12 + V2 d22 )
Thus
2
 m
2
 6  (0.12 m)
s


V3 =
 m 

 m
(cos17.2 )  4  (0.1m)2 +  6  (0.12 m)2 
 s 
 s 

and
m
V3 = 4.29
s
To determine the loss of available energy associated with the flow through this control
volume by applying the energy equation
∨ p V2

∂
e
ρ
d
V
+
+
+
+
u
gz

ρ V ⋅ nˆ dA = Qnet +Wshaft we obtain
CV  ρ 2
∂t CV
in
net in


 ∨ V2    ∨ V2    ∨ V2  
−  u1 + 1  m1 −  u2 + 2  m2 +  u3 + 3  m3 = 0 (4)



2 
2 
2 



Also, the conservation of mass equation, Eq. (3), can also be written as



− m1 − m2 + m3 = 0
(5)
Combining Eqs. (4) and (5), and assuming Q = 0 , we obtain
  ∨
∨ 
  ∨
∨ 
  V 2 −V 2 
  V 2 −V 2 
3
3
+ m2  2
m1  u3 − u1  + m2  u3 − u2  = m1  1





2
2








(6)
The left-hand side of Eq. (6) represents the rate of available energy loss in this fluid. Thus
rate of available energy loss is
 V 2 −V32 
 V 2 −V32 
rate of loss = ρV1A1  1
+ ρV2 A2  2





2
2




or
rate of loss =
ρπ  2  V12 −V32 
d1 V1 
4 

2
 V 2 −V32  
2
 + d2V2  2
 
2


 
Thus
2

  m 2 
m  

  4  −  4.29   
s   
(0.10 m)2  4 m    s  


2
 
 N ⋅s  
2
 s 
kg 


 
 
 999 3  (3.14) 1
kg ⋅ m  

 

m 

rate of loss =


2
4
  m 2 

m  
− 4.29   
 6

m    s  
s  

2
+(0.12 m)  6 

2

 s 




 

and
rate of loss = 558
N⋅m
s
Problem 5.129
Water flows vertically upward in a circular cross-sectional pipe. At section (1), the velocity
profile over the cross-sectional area is uniform. At section (2), the velocity profile is
17
 R−r 
V = wc 
 K
 R 
where V = local velocity vector, wc = centerline velocity in the axial direction, R = pipe
inside radius, and r = radius from pipe axis. Develop an expression for the loss in available
energy between sections (1) and (2).
Solution 5.129
Section (2)
Section (1)
Flow
For determining loss, we use the energy equation for nonuniform flows,
pout
ρ
2
+
loss =
α outV out
2
p1 − p2
ρ
+ gzout =
2
+
pin
ρ
2
+
α inV in
2
+ gzin + wshaft − loss . Thus,
net in
2
α1V 1 − α 2V 2
2
+ g ( z1 − z2 ) (1)
From conservation of mass Q = AV
1 1 = A2V 2 , we have
V1 =V 2
V2
A 2 ρ V ⋅ ndA
for the kinetic energy coefficient, α , we have
Also, with α =
 V2
m
2
α1 = 1.0
Since the velocity profile at section (1) is uniform. At section (2), we solve
V2
A 2 ρ V ⋅ ndA
α=
and obtain
 V2
m
2
α 2 = 1.06
Thus, Eq. (1) yields
loss =
p1 − p2
ρ
2
V1
− 0.06
+ g ( z1 − z2 )
2
Problem 5.130
Calculate the kinetic energy correction factor for each of the following velocity profiles for
a circular pipe:
r

(a) u = umax 1 −  ;
 R

r2 
(b) u = umax  1 − 2  ;
 R 


1
r 9

(c) u = umax  1 −  .
 R
R is the pipe radius and r is the radial coordinate.
Solution 5.130
The kinetic energy correction factor (called the kinetic energy coefficient in the text) is
 u2 
 dA
2
 
u2
m
2
 ρ u 
α=
For incompressible flow in a circular pipe, this can be simplified to
R
αn =
2  u3rdr
0
u 3R 2
r r
2  u3   d  
R R
=
u3
1
r n

The velocity profiles of part (a) and part (c), both fit the general format u = umax  1 −  .
 R
We shall first find α in terms of n .
3
1
r n  r   r 

2  u max  1 −    d  
 R R R
0
1
3
αn =
3
1


R
2
π
r
n



1−
u
rdr 
 π R 2  max  R 



0


3
1
r n  r   r 

u max  1 −    d  
R R R
0

3
=
1
1

r
n  r   r 


3

u max   1 −    d  
  R   R   R 
 0

3
=
3
(1 − z ) n ( z )dz
0
1

1
  (1 − z ) n ( z )dz 



0
3
Evaluating the integrals between 0 and 1 (with the help of mathematical software) gives,
after a bit of algebra
αn =
For n = 1 α n =
(1 + 1)3 (2 × 1 + 1)3
( n + 1)3 (2 n + 1)3
4 n 4 ( n + 3)( n + 2)
= 2.70 n = 9 α n =
(9 + 1)3 (2 × 9 + 1)3
4 × 14 (1 + 3)(2 × 1 + 3)
4 × 9 4 (9 + 3)(2 × 9 + 3)
The calculation for velocity profile (b) is given in Example 5.26. α = 2.0
= 1.037
Problem 5.131
The cross-sectional area of a rectangular duct is divided into 16 equal rectangular areas, as
shown in the figure below. The axial fluid velocity measured in feet per second in each
smaller area is shown. Estimate the kinetic energy correction factor.
20.0 in.
3.0
3.4
3.6
3.1
3.7
4.0
3.9
3.8
3.9
4.6
4.5
4.2
3.7
4.4
4.3
3.9
16.0 in.
Velocities in ft/s
Solution 5.131
The kinetic energy correction factor is
α=
1 3
u dA
A A
3
uave
=
1 3
u dA
A A
1

  udA
 A A

3
=
A2  u3dA
A


  udA
 A

The numerator integral is approximated as
16
 u dA ≈ 
3
j =1
A
 3.73 + 4.43 + 4.33 + 3.93 + 3.93 + 4.63 + 4.53 + 4.23   ft 3  5  4 
3
u j Aj = 
ft  ft 

3
3
3
3
3
3
3
3   

 +3.7 + 4.0 + 3.9 + 3.8 + 3.0 + 3.4 + 3.6 + 3.1   s   12
= 134.5
ft5
s3
The denominator integral is approximated as
16
 3.7 + 4.4 + 4.3 + 3.9 + 3.9 + 4.6 + 4.5 + 4.2   ft  5  4 
udA
≈
 u j Aj =  +3.7 + 4.0 + 3.9 + 3.8 + 3.0 + 3.4 + 3.6 + 3.1 s 
 ft  ft 

12  12 


1
j
=
A
= 8.611
ft3
s
Then
α≈
2
ft5 
 20 16 2  
ft
134.5
×


 
s3 
 12 12
 
α ≈ 1.04
3

ft3 
 8.611 
s 

 12 
Problem 5.132
A small fan moves air at a mass flowrate of 0.004 lbm/s . Upstream of the fan, the pipe
diameter is 2.5in. , the flow is laminar, the velocity distribution is parabolic, and the kinetic
energy coefficient, α1, is equal to 2.0. Downstream of the fan, the pipe diameter is 1in. , the
flow is turbulent, the velocity profile is quite flat, and the kinetic energy coefficient, α 2 , is
equal to 1.08. If the rise in static pressure across the fan is 0.015 psi and the fan shaft draws
0.00024 hp, compare the value of loss calculated: (a) assuming uniform velocity
distributions, (b) considering actual velocity distributions.
Solution 5.132
(a) For uniform velocity distributions upstream of the fan, the equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss is applicable. Thus,
2
2
ρ
ρ
net in
loss =
pin − pout
ρ
2
0 for air
Vin2 −Vout


zout  + wshaft
+
+ g  zin −
2


net in
(1)
We obtain the shaft work, wshaft from the given shaft power, Wshaft , with
net in
net in

ft ⋅ lb 
(0.00024 hp)  550
s ⋅ hp 
ft ⋅ lb
net in

=
= 33

lbm
lbm
0.004
m
s
W shaft
wshaft =
net in

For Vin and Vout , we use the equation m = ρ AV . Thus,
Vin =


m
=
ρ Ain
m
=
π Din2
ρ
4
lbm  
in.2 

0.004
144



s  

ft 2 
lbm   π (2.5 in.)2 

−3 slug  
×
2.38
10
32.2




slug  
4
ft3  


= 1.53
ft
s
and
Vout =


m
=
ρ Aout
m
ρ
2
π Dout
4
=
lbm  
in.2 

0.004
144



s  

ft 2 
lbm   π (1in.)2 

−3 slug  
×
2.38
10
32.2




slug  
4
ft3  


= 9.57
ft
s
Now from Eq. (1), we obtain
2
2


in.2 
ft  
ft   

( −0.015 psi) 144 2 
−
1.53
9.57




 
 

ft
lb
1
s
s


 
  1
+ 
loss =



ft
2
lbm  

−3 slug  
slug ⋅ 2   32.2 lbm 
32.2



 2.38 × 10


s 
slug  
slug 
ft3  

 
+ 33
ft ⋅ lb
lbm
or
loss = 3.43
ft ⋅ lb
lbm
(b) For nonuniform velocity distributions upstream and downstream of the fan, the
equation
pout
ρ
2
+
α outV out
2
+ gzout =
pin
ρ
2
+
α inV in
2
+ gzin + wshaft − loss is applicable.
net in
Thus,
loss =
pin − pout
ρ
2
+
2
αinV in − α outV out
2
0 for air


zout  + wshaft
+ g  zin −


net in
or
2
2
 
ft 
ft   


(2)
1.53
(1.08)
9.57
 




ft ⋅ lb  
lb 
1
ft ⋅ lb
s
s 

+
−
+ 33
loss = −28.18
1



ft
lbm
2
2
lbm
lbm 



  slug ⋅ 2   32.2

s 
slug 

 
and
loss = 3.36
ft ⋅ lb
lbm
Problem 5.133
Air enters a radial blower with zero angular momentum. It leaves with an absolute
tangential velocity, Vθ , of 200 ft/s. The rotor blade speed at rotor exit is 170 ft/s. If the
stagnation pressure rise across the rotor is 0.4 psi , calculate the loss of available energy
across the rotor and the rotor efficiency.
Solution 5.133
To determine the loss of available energy across the rotor we use the energy equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss to obtain
2
2
ρ
ρ
net in
loss =
pin − pout
ρ
2
0, neglect
Vin2 −Vout


+
+ g  zin −
zout  + wshaft
2


net in
or
loss =
p0,in − p0,out
ρ
+ wshaft
(1)
net in
The shaft work in, wshaft can be obtained with the moment-of-momentum work equation
net in
wshaft = −( ±UinVθ in ) + ( ±U outVθ out ).
Thus,
wshaft = U outVθ out
(2)
net in
Combining Eqs. (1) and (2) leads to
loss =
p0,in − p0,out
ρ
+ U outVθ out
or

in.2 


(0.4 psi)  144 2 




ft
ft
ft
lb






+ 170   200  1
loss = −

s 
s   slug ⋅ ft 

 



lb 

s2 
−3 slug 

×
2.38
10
1




ft3   slug ⋅ ft 



s2 

and
loss = 9800
ft ⋅ lb
ft ⋅ lb
1
ft ⋅ lb
= 9800
= 305
slug
slug 
lbm
lbm 
 32.174 slug 


Then we calculate rotor efficiency from
wshaft − loss
rotor efficiency =
net in
wshaft
net in
=
U outVθ ,out − loss
U outVθ ,out


ft  
ft  
lb 
ft ⋅ lb

 − 9800
 170   200  1
s 
s   slug ⋅ ft 
slug


2 
s 

= 0.71
rotor efficiency =


ft  
ft  
lb 


 170   200  1
s 
s   slug ⋅ ft 



s2 

Problem 5.134
Water enters a pump impeller radially. It leaves the impeller with a tangential component of
absolute velocity of 10 m/s . The impeller exit diameter is 60 mm , and the impeller speed is
1800 rpm . If the stagnation pressure rise across the impeller is 45kPa , determine the loss of
available energy across the impeller and the hydraulic efficiency of the pump.
Solution 5.134
 2.0w 2 1.0w 2

2
1
The equation p1 − p2 = ρ 
−
+ g ( z2 − z1 ) + loss  is applicable to solving this
2
 2

problem. Using the equation above, we obtain
loss = U 2Vθ ,2 −
(actual total pressure rise across inpeller)
ρ
However,
rev   rad 

(60 mm) 1800
  2π

m
min  
rev 

U 2 = r2ω =
= 5.66
mm  
s 
s

(2)  1000
  60

m   min 

Thus,



m  m 
N  
1

3 N 
loss =  5.66  10  1
 −  45 ×10

2
s   s   kg ⋅ m  
m   999 kg 




2 
s 


m3 
loss = 11.6
N⋅m
kg
2
4
6
r
r r 
From α 2 = 2  1 − 3   + 3   −    r dr , we obtain
R 0 
R
 R   R  
16
R
(actual total pressure rise across impeller)
η=
ρ
U 2Vθ ,2
or

3 N 

  45 × 10
m2  

  999 kg  

 
m3  
= 0.796
η=


m  m 
N 


 5.66  10  1
s   s   kg ⋅ m 



s2 

Problem 5.135
Water enters an axial-flow turbine rotor with an absolute velocity tangential component,
Vθ , of 15ft/s . The corresponding blade velocity, U , is 50ft/s . The water leaves the rotor
blade row with no angular momentum. If the stagnation pressure drop across the turbine is
12 psi , determine the hydraulic efficiency of the turbine
Solution 5.135
To determine the efficiency of the turbine, we use
η=
actual work out
(1)
actual work out + loss
The actual work out, wshaft , is obtained with the moment-of-momentum work equation
net out
wshaft = −( ±UinVθ in ) + ( ±U outVθ out )
Thus,
wshaft
net out
= −wshaft = UinVθ ,in (2)
net in
To determine the loss of available energy across the rotor, we use the energy equation
2
pout Vout
p
V2
+
+ gzout = in + in + gzin + wshaft − loss to obtain
ρ
ρ
2
2
net in
loss =
pin − pout
ρ
+
2
neglect
Vin2 −Vout


+ g  zin −
zout  + wshaft (3)
2


net in
Combining Eqs. (2) and (3), we obtain
loss =
pin − pout
ρ
+ UinVθ in
(4)
Combining Eqs. (1), (2), and (4), we obtain




ft
ft
lb




 50 15  1
 s  s   slug ⋅ ft 


U inVθ ,in
UinVθ ,in
s2 

η=
=
=
UinVθ ,in + loss p0,in − p0,out

in.2 
12 psi ) 144 2 
(

ρ
ft 

slugs 

1.94 3 
ft 

and
η = 0.842
Problem 5.136
An inward flow radial turbine (see the figure below) involves a nozzle angle, α1, of 60° and
an inlet rotor tip speed, U1, of 30ft/s . The ratio of rotor inlet to outlet diameters is 2.0. The
radial component of velocity remains constant at 20ft/s through the rotor, and the flow
leaving the rotor at section (2) is without angular momentum. If the flowing fluid is water
and the stagnation pressure drop across the rotor is 16 psi , determine the loss of available
energy across the rotor and the hydraulic efficiency involved.
60°
Vr1 =
20 ft/s
r1
r2
2
U1 =
30 ft/s
1
Solution 5.136
Fixed
control
volume
60°
VR1 =
20 ft/s
r1
r2
(2)
U1 =
30 ft/s
(1)
Since a turbine is involved in this problem, wshaft = −wshaft and from the equation
net in
p2
ρ
+
V22
loss =
2
+ gz2 =
p1
ρ
+
V12
2
net out
+ gz1 + wshaft − loss , we can conclude that
net in
(stagnation pressure drop across rotor)
ρ
− wshaft
net out
However, from the equation, wshaft = −( ±UinVθ in ) + ( ±U outVθ out ), we see that
wshaft = wshaft = −U1Vθ ,1 = −wshaft
net in
net out
and thus
loss =
(stagnation pressure drop across rotor)
ρ
−U1Vθ ,1 (1)
To determine the value of Vθ ,1 , we examine the velocity triangle for the flow entering the
rotor that is sketched below.
60°
V1
Vr1
W1
U1
Vθ ,1
From the velocity triangle, we obtain
Vθ ,1 = Vr1 tan 60
or
ft
 ft 
Vθ ,1 =  20  tan60 = 34.64
s
 s
From Eq. (1), we obtain
lb  
in.2 



16
144



ft 2   ft 
 in.2  
ft  
lb 
−  30  34.64  1
loss =

slugs 
s 
s   slug ⋅ ft 




1.94 3 
s2 

ft 

loss = 148
From
ft ⋅ lb
slug
pout
ρ
+
2
Vout
p
V2
+ gzout = in + in + gzin + wshaft − loss , we can conclude that
2
2
ρ
net in
wshaft + loss =
(stagnation pressure drop across the rotor)
ρ
net out
or in other words, the stagnation pressure drops across the rotor result in shaft work and
loss of available energy.
Thus a meaningful efficiency is
wshaft
η=
or
net out
 stagnation pressure drop across the rotor 


ρ





ft 
lb 
 ft 

 30  34.64  1
s   slug ⋅ ft 
 s 


s2  = 0.875

η=
lb  
in.2 

16
144




ft 2 
 in.2  
slugs 

1.94 3 
ft 

Problem 6.1
The velocity in a certain two-dimensional flow field is given by the equation V = 2 xtˆi – 2 ytˆj
ft
where the velocity is in when x, y, and t are in feet and seconds, respectively. Determine
s
expressions for the local and convective components of acceleration in the x and y directions. What is the magnitude and direction of the velocity and the acceleration at the point
x = y = 2 ft at the time t = 0?
Solution 6.1
From the expression for velocity, u = 2xt and v = −2 yt .
Since
ax =
∂u
∂u
∂u
+u +v
∂t
∂x
∂y
Then
ax (local) =
∂u
= 2x
∂t
And
ax (conv) = u
∂u
∂u
+v
= (2xt )(2t ) + ( −2 yt )(0) = 4xt 2
∂x
∂y
Similarly,
ay =
∂v
∂v
∂v
+u +v
∂t
∂x
∂y
And
ay (local) =
∂v
= −2 y
∂t
ay (conv) = u
∂v
∂v
+v
= (2xt )(0) + ( −2 yt )( −2t ) = 4 yt 2
∂x
∂y
At x = y = 2 ft and t = 0
u = 2(2)(0) = 0v = −2(2)(0) = 0
So that V = 0
ft
and ax = 2 x + 4 xt 2 = 2(2) + 4(2)(0) = 4 2
s
ft
a y = −2 y + 4 yt 2 = −2(2) + 4(2)(0) = −4 2
s
ft
ft
Thus, a = 4ˆi – ˆj 2 with | a | = (4)2 + ( −4)2 = 5.66 2
s
s
Problem 6.2
The velocity in a certain flow field is given by the equation V = xˆi x 2 zˆj + yzkˆ . Determine
the expressions for the three rectangular components of acceleration.
Solution 6.2
From expression for velocity, u = x, v = x2z, w = yz
Since
ax =
∂u
∂u
∂u
∂u
+u +v +w
∂t
∂x
∂y
∂z
Then
ax = 0 + (x)(1) + (x2 z )(0) + ( yz )(0) = x
Similarly,
ay =
∂v
∂v
∂v
∂v
+u +v +w
∂t
∂x
∂y
∂z
And
ay = 0 + ( x )(2 xz ) + ( x 2 z )(0) + ( yz )( x 2 ) = 2x 2 z + x 2 yz
Also,
az =
∂w
∂w
∂w
∂w
+u
+v
+w
∂t
∂x
∂y
∂z
So that
az = 0 + ( x )(0) + ( x 2 z )( z ) + ( yz )( y ) = x 2 z 2 + y2 z
Problem 6.3
The flow in the plane two-dimensional channel shown in the figure below has x- and ycomponents of velocity given by
2
 x   y  
u = uo 1 +  1 −   

   Y  
 y3  x 2 y 
v = uo  2 1 +  − 

 Yo 

Calculate the linear acceleration, rotation, vorticity, rate of volumetric strain, and rate of
shear deformation for the flow.
y
Y0 = 1.0 m
Y (x)
u (x, y )
+
Yℓ = 0.5 m
x
ℓ = 5.0 m
Solution 6.3
There are two linear accelerations, a x and ay . Equations (5.1) and (5.2) give
ax =
Du ∂u
∂u
∂u
= + u +v ;
Dt ∂t
∂x
∂y
ay =
Dv ∂v
∂v
∂v
= +u +v .
Dt ∂t
∂x
∂y
By inspection,
∂u ∂v
=
= 0.
∂t ∂t
Before continuing, we substitute
Y=
Yo
1 + ( x
) 
into the equation for u, which gives
2
2
 x  y  x  
u = uo 1 +  1 − 2 1 +  

  Yo 
 
 x  y2  x 3 
= uo 1 +  − 2 1 +   .
 Yo 
 

Then
 1 3 y 2  x 2 
∂u
= uo  −
1 +   ;
∂x
Yo2 
 

 2 y  x 3 
∂u
= uo  − 2 1 +   ;
∂y
 
 Yo 
 2 y3  x  
∂v
= uo  2 2 1 +   ;
∂x
 
 Yo 
 3 y 2  x 2 1 
∂v
= uo  2 1 +  −  .
∂y

 Yo 

Substituting and simplifying, we obtain
2
4
3
5
uo2  x   y   x   y   x  
 1 +  − 2    1 +  +    1 +  
ax =

  Yo  
  Yo  
 


and
2
4
uo2 
y3  x 
y5  x  
ay = 2  y − 2 2 1 +  + 4 1 +   .
Yo 
 Yo 
 

The angular velocity is
1  ∂v
∂u 
ωz =  −  .
2  ∂x ∂y 
Substituting for ∂v ∂x and ∂u dy , we have
2
yuo  x   y2  x  
ωz = 2  1 +   2 +  1 +   .
Yo 
 

 
The vorticity is
ξ z = 2ω z ,
so
ξz =
2
2 yuo  x   y2  x  
1
1
+
+
+

 2 
 .
Yo2 
 

 
The rate of volumetric strain is
1 d (δ V )  ∂u ∂v 
=  + .
δV
dt
 ∂x ∂y 
Substituting for ∂ u ∂ x and ∂v dy, we have
1 d (δ V )
= 0.
δV
dt
The rate of shear deformation is
⋅
1  ∂u
∂v 
φ =  + .
2  ∂y ∂x 
Substituting for ∂u ∂y and ∂v ∂x , we obtain
2
yuo  x    x  y2 
φ = 2 1 +   − 1 +  + 2  .
Yo 
  


⋅
Comment It would be informative if we could observe the rotation, stretching, and deformation of a fluid particle as it moved through a flow field. Even though we can mathematically separate each quantity, the rotation, stretching, and deformation occur simultaneously
and one cannot be isolated from the others. Calculating a few numerical values of ∂u ∂y
and ∂v ∂x and comparing them would be instructive for you.
Problem 6.4
The three components of velocity in a flow field are given by
u = x2 + y2 + z 2
v = xy + yz + z 2
z2
w = −3xz − + 4
2
(a) Determine the volumetric dilatation rate and interpret the results. (b) Determine an
expression for the rotation vector. Is this an irrotational flow field?
Solution 6.4
(a) Volumetric dilatation rate =
∂u ∂v ∂w
+ +
∂x ∂y ∂z
Thus, for velocity components given volumetric dilatation rate = 2x + ( x + z ) +
( −3x − z ) = 0
This result indicates that there is no change in the volume of a fluid element as it moves
from one location to another.
(b) For the axes z, x, y, the velocity components given:
1  ∂v
∂u 
1
1  ∂w
∂v 
1
1  ∂u
∂w 
1
y
ωz =  −  = ( y − 2 y) =
2  ∂x ∂y  2
2
y

ωx =  −  = [0 − ( y + 2 z )] = −  + z 
2  ∂y ∂z  2
2

5z
ωy =  −
 = [2 z − ( −3 z )] =
2  ∂z ∂x  2
2
y
 5z ˆ y ˆ
j– k
Thus, ω = −  + z  ˆi
2
2
2

Since ω is not zero everywhere the flow field is not irrotational. No.
Problem 6.5
Determine an expression for the vorticity of the flow field described by V = − xy3 ˆ + y 4 ˆ
Is the flow irrotational?
Solution 6.5
=2
From expression for velocity, u = − xy3, v = y 4 , and w = 0, and with
1  ∂w
∂v 
1  ∂u
∂w 
1  ∂v
∂u 
ωx =  − 
2  ∂y ∂z 
ωy =  −

2  ∂z ∂x 
ωz =  − 
2  ∂x ∂y 
It follows that
1
3
ω x = 0, ω y = 0, and ωz = [0 − ( −3xy2 )] = xy2
2
2
Thus,

3
 
ς = 2ω = 2( ωxˆi + ω yˆj + ωz kˆ ) = 2 (0)ˆi +(0)ˆj +  xy2  kˆ  = 3xy2 kˆ
2
 

Since is not zero everywhere the flow is not irrotational. No.
Problem 6.6
According to Eq. (6.134), the x -velocity in fully developed laminar flow between parallel
plates is given by
u=
(
1  dp  2 2
  y −h
2 μ  dx 
)
The y-velocity is v = 0. Determine the volumetric strain rate, the vorticity, and the rate of
angular deformation. What is the shear stress at the plate surface?
Solution 6.6
The given velocity distribution is u =
where μ, h, and
(
)
1  dp  2
2
  y − h and v = 0
2 μ  dx 
dp
are constants.
dx
The required derivatives are
∂v
∂u 1  dp 
∂u
∂v
=0
=   ( y );
= 0;
= 0;
∂y
∂y μ  dx 
∂x
∂x
The volumetric strain rate is
1  d (δ ∀ )  ∂u ∂v
+
=0+0 =0

=
δ ∀  dt  ∂x ∂y
The vorticity is ς = ς z =
∂v ∂u
1  dp 
y  dp 
−
= 0 −   ( y) = −  
∂x ∂y
μ  dx 
μ  dx 
The rate of angular deformation is γ xy =
∂v ∂u
1  dp 
y  dp 
+
= 0 +   ( y) =  
∂x ∂y
μ  dx 
μ  dx 
And, finally, the shear stress at y = 0 is τ y=h = μγ xy, y=h = μ
y  dp  
 dp 
= h 
 
μ  dx   y=h
 dx 
Problem 6.7
For a certain incompressible, two-dimensional flow field the velocity component in the
y direction is given by the equation
v = 3xy + x 2 y
Determine the velocity component in the x direction so that the volumetric dilatation rate is
zero.
Solution 6.7
For zero volumetric rate in a two-dimensional flow,
∂u ∂v
+
=0
∂x ∂y
Since
(1)
∂v
= 3x + x2
∂y
Then from Eq. (1)
∂u
= −3x − x 2
∂x
(2)
Equation (2) can be integrated with respect to x to obtain
 du = − 3xdx −  x dx + f ( y)
2
Or
3
x3
u = − x2 −
f ( y)
2
3
where f ( y ) is an undetermined function of y .
Problem 6.8
An incompressible viscous fluid is placed between two large parallel plates as shown in the
figure below. The bottom plate is fixed and the upper plate moves with a constant velocity,
U . For these conditions, the velocity distribution between the plates is linear and can be
expressed as
y
u =U
b
Determine: (a) the volumetric dilatation rate, (b) the rotation vector, (c) the vorticity, and
(d) the rate of angular deformation.
U
Moving
plate
u
b
y
Fixed
plate
Solution 6.8
∂u ∂v ∂w
+ +
= 0+0+0 = 0
∂x ∂y ∂z
(b) For the velocity distribution given,
= ω z kˆ
(a) Volumetric dilation rate =
And
1  ∂v
∂u 
=−
U ˆ
k
2b
1
∂ 
y 
U
ωz =  −  =  0 − U   = −
2  ∂x ∂y  2 
∂y  b  
2b
Thus,
(c) ς = 2ω = −
(d) γ =
U ˆ
k
b
∂v ∂u
∂  y U
= 0 + U  =
+
∂x ∂y
∂y  b  2b
Problem 6.9
A viscous fluid is contained in the space between concentric cylinders. The inner wall is
fixed, and the outer wall rotates with an angular velocity ω . (See the figure (a) below)
Assume that the velocity distribution in the gap is linear as illustrated in the figure (b) below. For the small rectangular element shown in the figure (b) below, determine the rate of
change of the right angle γ due to the fluid motion. Express your answer in terms of r0, ri,
and ω.
ω
ro ω
u
ri
ro
γ
y
ro – ri
x
(a)
(b)
Solution 6.9
From
γ=
∂v ∂u
+
∂x ∂y
For the linear distribution
u=−
r0ω
y
r0 − ri
So that
rω
∂u
=− 0
r0 − ri
∂y
And since v = 0
γ =−
r0ω
r0 − ri
The negative sign indicates that the original right angle is increasing.
Problem 6.10
Air is delivered through a constant-diameter duct by a fan. The air is inviscid, so the fluid
velocity profile is “flat” across each cross section. During the fan start-up, the following
velocities were measured at the time t and axial positions x :
x=0
x = 10 m
x = 20 m
t=0s
V = 0 m/s
V = 0 m/s
V = 0 m/s
t = 1.0 s
V = 1.00 m/s V = 1.20 m/s V = 1.40 m/s
t = 2.0 s
V = 1.70 m/s V = 1.80 m/s V = 1.90 m/s
t = 3.0 s
V = 2.10 m/s V = 2.15 m/s V = 2.20 m/s
Calculate the local acceleration, the convective acceleration, and the total acceleration at
t = 1.0 s and x = 10 m. What is the local acceleration after the fan has reached a steady air
flow rate?
Solution 6.10
The local acceleration is ax,L =
∂u ∂V
=
∂t
∂t
The convective acceleration is ax,C = u
The total acceleration is ax =
∂u
∂V
=V
∂x
∂x
∂u
∂u ∂V
∂V
+u
=
+V
∂t
∂x ∂t
∂x
For the discrete data, the derivatives will be approximated by central finite differences
aL =
Vt =2,x =10 −Vt =0,x =10
∂V 
≈
=
∂t t =1,x =10 tt =2,x =10 − tt =0,x =10
m
m
−0
s
s = 0.90 m
2s − 0s
s2
1.80
m
m
1.40 − 1.00
Vt =1,x =20 −Vt =1,x =0
m
∂V 
s
s = 0.024 m
≈ Vt =1,x =10
= 1.20 ×
aC = V

∂x t =1,x =10
s
20 m − 0 m
xt =1,x =20 − xt =1,x =0
s2
m
m
m
aT = aL + aC ≈ 0.90 2 + 0.024 2 = 0.924 2
s
s
s
For steady flow
aL = 0 by definition
Problem 6.11
For a certain incompressible flow field, it is suggested that the velocity components are given by the equations
u = 2xy v = − x 2 y w = 0
Is this a physically possible flow field? Explain.
Solution 6.11
Any physically possible incompressible flow field must satisfy conservation of mass as expressed by the relationship
∂u ∂v ∂w
+ +
=0
∂x ∂y ∂z
(1)
For the velocity distribution given,
∂v
∂u
∂w
= −x2
= 2y
=0
∂y
∂x
∂z
Substitution into Eq. (1) shows that
2 y − x2 + 0 ≠ 0
Thus, this is not a physically possible incompressible flow field. No.
Problem 6.12
The velocity components of an incompressible, two-dimensional velocity field are given by
the equations
u = y 2 − x(1 + x )
v = y(2x + 1)
Show that the flow is irrotational and satisfies conservation of mass.
Solution 6.12
If the two-dimensional flow is irrotational,
1  ∂v
∂u 
ωz =  −  = 0
2  ∂x ∂y 
For the velocity distribution given,
∂v
= 2y
∂x
∂u
= 2y
∂y
Thus,
1
2
ωz = (2 y − 2 y ) = 0
And the flow is irrotational.
To satisfy conservation of mass,
∂u ∂v
+
=0
∂x ∂y
Since,
∂v
∂u
= 2x + 1
= −1 − 2x and
∂y
∂x
Then
−1− 2x + 2x +1 = 0
And
Conservation of mass is satisfied.
Problem 6.13
Consider the two-dimensional channel flow of Problem 6.3. Show that the given velocities
satisfy conservation of mass in both differential and control volume forms.
Solution 6.13
The control volume form of conservation of mass is
∂
ˆ dA = 0
cv ρ d ∀ +  ρ V • n
∂t 
A fixed, non-deforming control volume is shown in the figure
y ( ˆj)
Control volume
n̂
n̂
u
u
ˆ
x (i)
ℓ
As the flow is steady
 ρV • nˆ dA = 0.
∂
 ρ d ∀ = 0 and because ρ is constant, we need to show
∂t cv
For the control volume shown V • nˆ = 0 along the walls of the channel, because u = v = 0 at
the wall. The flow is two-directional, so
V = uiˆ + vjˆ.
At the outlet plane, the unit outward normal vector is +iˆ. At the inlet plane, the unit outward normal vector is −iˆ. Therefore
 V • nˆ dA = Aout ( uiˆ + vjˆ ) ⋅ (iˆ ) dA + Ain ( uiˆ + vjˆ ) ⋅ ( −iˆ ) dA
=
Aout
u dA − 
Ain
u dA.
Substituting for u at the exit,
2


  y  
Aout u dA = Aout uo 1 +  1 − Y   dA


2
+Y 
 y 
= Wuo  2 1 −    dy,
−Y
 Y  


Y is the magnitude of the channel half-height at x = and W is the width of the channel.
Integrating,
+Y

8WuoY
y3 
=
−
=
u
dA
2
Wu
y
.


o
Aout
3
3Y 2  −Y

As Y = 0.5 m and uo = 1.0 m s ,
4
Aout u dA = 3 W .
Similarly,
  y 2 
Ain u dA = Ain uo (1) 1 − Yo   dA


2
+Yo 
 y 
= Wuo  1 −    dy,
−Yo 
Y 
  o 
where Y o is the magnitude of the channel half-height at x = 0. Integrating,
+Y
o

8Wu Y
y3 
Ain u dA = Wuo  y − 3Y 2  = 3o o .
o  −Y

o
As Yo = 1.0 m and uo = 1.0 m s ,
4
Ain u dA = 3 W .
Then
4
4
 ρ V • nˆ dA = 3 W − 3 W = 0.
Thus
∂
ˆ dA = 0
cv ρ d ∀ +  ρ V • n
∂t 
Conservation of mass in differential form is expressed by the continuity equation. For incompressible two-dimensional flow
∂u ∂v
+
=0
∂x ∂y
Before continuing, we substitute
Y=
Yo
1 + ( x
) 
into the equation for u, which gives
2
2
 x  y  x  
u = uo 1 +  1 − 2 1 +  

  Yo 
 
 x  y2  x 3 
= uo 1 +  − 2 1 +   .
 Yo 
 

Then
 1 3 y 2  x 2 
∂u
= uo  −
1 +   ;
∂x
 
Yo2 

and
 3 y 2  x 2 1 
∂v
= uo  2 1 +  −  .
dy

 Yo 

Adding terms
∂u ∂v
+
=0
∂x ∂y
Showing that the continuity equation in differential form is satisfied.
Problem 6.14
The x-velocity profile in a certain laminar boundary layer is approximated as follows
π y 
u = U 0 sin 

 2 0.1 x 
Determine the y-velocity, v( x, y ).
Solution 6.14
Assuming two-dimensional incompressible flow, the continuity equation is
∂u ∂v
+
=0
∂x ∂y
So
∂v
∂u
∂ 
π y  
 πy 
=−
= − U 0 sin 
 = 5πU 0 y cos 


∂y
∂x
∂x 
 2 0.1 x  
 0.1 x 
Integrating

 πy 
 πy 
 π y 
2
2
v = 5πU 0  y cos 
 + 0.1π xy sin 
  + constant( x )
 dy = 5πU 0  0.01π x cos 
 0.1 x 
 0.1 x 
 0.1 x  

where the “constant” is at most a function of x .
NOTE: For an actual boundary layer, v = 0 at y = 0 so the constant = 0.
Problem 6.15
m2
For each of the following stream functions, with units of
, determine the magnitude
s
and the angle the velocity vector makes with the x axis at x = 1m, y = 2 m. Locate any
stagnation points in the flow field.
(a) ψ = xy
(b) ψ = −2x 2 + y
Solution 6.15
From the definition of the stream function,
u=
∂ψ
∂ψ
v=
∂y
∂x
(a) For ψ = xy,
u=
∂ψ
∂ψ
=x v=−
=y
∂y
∂x
At x = 1m, y = 2 m, it follows that u = 1
m
m
and v = −2
s
s
Thus,
| V | = u 2 + v 2 = (1 m)2 + ( −2 m)2 = 2.24
m
s
x
θ
–2
1
V
tan θ =
−2
1
θ = −63.4°
Since u = 0 at x = 0 and v = 0 at y = 0, a stagnation point occurs at x = y = 0
(b) For ψ = −2x 2 + y ,
u=
∂ψ
m
=1
∂y
s
v=−
∂ψ
= 4x
∂x
At x = 1m, y = 2 m, it follows that u = 1
m
m
and v = 4
s
s
Thus,
2
2
m
 m  m
| V | = u + v = 1  +  4  = 4.12
s
 s   s 
2
2
V
4
1
θ
x
tan θ =
4
1
Since u ≠ 0, there are no stagnation points.
θ = 76.0°
Problem 6.16
The stream function for an incompressible, two-dimensional flow field is
ψ = ay − by3
where a and b are constants. Is this an irrotational flow? Explain.
Solution 6.16
For the flow to be irrotational,
1  ∂v
∂u 
ωz =  −  = 0
2  ∂x ∂y 
And for the stream function given,
∂ψ
= a − 3by2
u=
∂y
∂ψ
v=−
=0
∂x
∂u
∂v
= −6by
Thus,
=0
∂y
∂x
So that
1
ωz = [0 − ( −6by )] = 3by
2
Since ω z ≠ 0 flow is not irrotational (unless b = 0). No.
Problem 6.17
For a certain two-dimensional flow field
u=0
u =V
(a) What are the corresponding radial and tangential velocity components? (b) Determine
the corresponding stream function expressed in Cartesian coordinates and in cylindrical polar coordinates.
Solution 6.17
(a) At an arbitrary point P (see figure)
V
y
θ
vθ
vr
P
r
θ
x
Vr =V sinθ
Vθ = V cosθ
(b) Since
u=
∂ψ
=0
∂y
It follows that ψ is not a function of y and
v=−
∂ψ
=V
∂x
Integrating
ψ = −Vx + C
where C is an arbitrary constant.
Also, with x = r cosθ
ψ = –Vr cos θ + C
Check this result:
vθ = −
∂ψ
= −( −V cos θ ) = V cos θ
∂r
and
vr =
1 ∂ψ 1
= (Vr sin θ ) = V sin θ , which checks with part (a).
r ∂θ r
Problem 6.18
In a certain steady, two-dimensional flow field, the fluid density varies linearly with respect
to the coordinate x : that is, ρ = Ax where A is a constant. If the x component of velocity u is
given by the equation u = y, determine an expression for v .
Solution 6.18
For a variable density flow,
∂( ρ u ) ∂( ρv )
+
=0
∂x
∂y
With
ρ u = ( Ax )( y ) = Axy
If follows that
∂( ρ u )
= Ay
∂x
∂( ρv )
= −Ay
Thus,
∂y
(1)
Integrate Eq. (1) with respect to y to obtain
 d( ρv) = − Aydy + f1(x)
or
ρv = −
Ay2
+ f1 ( x )
2
With ρ = Ax
2
 1   Ay  f1( x )
v = −
+

 Ax   2  Ax
or
v=−
y2
+ f (x)
2x
where f ( x ) is an arbitrary function of x
Problem 6.19
In a two-dimensional, incompressible flow field, the x component of velocity is given by the
equation u = 2x. (a) Determine the corresponding equation for the y component of velocity
if v = 0 along the x axis. (b) For this flow field, what is the magnitude of the average velocity
of the fluid crossing the surface OA of the figure below? Assume that the velocities are in
feet per second when x and y are in feet.
y, ft
1.0
A
O
1.0
x, ft
Solution 6.19
y, ft
A
1.0
QOA
QAB
QOB
O
B
1.0
x, ft
(consider a unit thickness = 1ft)
(a) To satisfy the continuity equation
∂u ∂v
+
=0
∂x ∂y
Since
∂u
=2
∂x
It follows that
∂v
= −2
∂y
(1)
Integration of Eq. (1) with respect to y yields
v = −2 y + f ( x )
If v = 0 along x-axis ( y = 0) then f ( x ) = 0 so that v = −2 y
(b) To satisfy conservation of mass
QOA = QAB − QOB (see the figure above)
Along AB u = 2(1) = 2
ft
so that
s
ft3
 ft 
QAB = uAAB =  2  (1ft)(1ft) = 2
s
 s
Along OB v = 0 so that QOB = 0
ft3
Thus, QOA = QAB = 2
s
and
ft 3
QOA
s = 1.41 ft
VAV =
=
areaOA
s
2 ft 2
2
Problem 6.20
The stream function for an incompressible flow field is given by the equation
ψ = 3x 2 y − y3
m2
where the stream function has the units of
with x and y in meters. (a) Sketch the
s
streamline(s) passing through the origin. (b) Determine the rate of flow across the straight
path AB shown in the figure below.
y, m
1.0
B
A
0
1.0
x, m
Solution 6.20
(a) Lines of constant ψ are streamlines. For ψ = 3x 2 y − y3, the streamline passing
through the origin (x = 0, y = 0) has the value ψ = 0. Thus, the equation for the
streamlines through the origin is
0 = 3x 2 y − y3
or
y = ± 3x
A sketch of these streamlines is shown in the figure below.
ψ=0
y
ψ=0
√3
√3
1
1
ψ=0
ψ=0
x
(b) Q = ψ B −ψ A
At B x = 0, y = 1 m so that
ψ B = 3(0)2 (1) − (1)3 = −1
m3
(per unit width)
s
At A x = 1m, y = 0
so that
ψ A = 3(1)2 (0) − (0)3 = 0
Thus,
Q = ψ B = −1
m3
( per unit width )
s
The negative sign indicates that the flow is from right to left as we look from A to B
Problem 6.21
The stream function for an incompressible, two-dimensional flow field is
ψ = 3x 2 y + y
For this flow field, plot several streamlines.
Solution 6.21
The equation for a streamline is found by setting ψ constant in the equation for the
stream function. Thus, for the given stream function.
ψ = 3x 2 y + y
It follows that the equation of a streamline is
y=
ψ
1 + 3x 2
where various constant values can be assigned to ψ to obtain a family of streamlines. Tabulated results for ψ = 1,2,3, 4, and a plot showing the streamlines are given below.
x
−5.0
−4.5
−4.0
−3.5
−3.0
−2.5
−2.0
−1.5
−1.0
−0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
=1
y
0.0132
0.0162
0.0204
0.0265
0.0357
0.0506
0.0769
0.1290
0.2500
0.5714
1.0000
0.5714
0.2500
0.1290
0.0769
0.0506
0.0357
0.0265
0.0204
0.0162
0.0132
=2
y
0.0263
0.0324
0.0408
0.0530
0.0714
0.1013
0.1538
0.2581
0.5000
1.1429
2.0000
1.1429
0.5000
0.2581
0.1538
0.1013
0.0714
0.0530
0.0408
0.0324
0.0263
=3
y
0.0395
0.0486
0.0612
0.0795
0.1071
0.1519
0.2308
0.3871
0.7500
1.7143
3.0000
1.7143
0.7500
0.3871
0.2308
0.1519
0.1071
0.0795
0.0612
0.0486
0.0395
=4
y
0.0526
0.0648
0.0816
0.1060
0.1429
0.2025
0.3077
0.5161
1.0000
2.2857
4.0000
2.2857
1.0000
0.5161
0.3077
0.2025
0.1429
0.1060
0.0816
0.0648
0.0526
4.50
4.00
ψ=4
3.50
ψ=3
3.00
y
2.50
2.00
ψ=2
1.50
ψ=1
1.00
0.50
–5.0
–3.0
0.00
–1.0
1.0
x
3.0
5.0
Problem 6.22
Consider the incompressible, two-dimensional flow of a nonviscous fluid between the
boundaries shown in the figure below. The velocity potential for this flow field is
φ = x 2 − y2
(a) Determine the corresponding stream function. (b) What is the relationship between the
discharge, q (per unit width normal to plane of paper) passing between the walls and the
coordinates x i , y i of any point on the curved wall? Neglect body forces.
y
q
(xi, yi)
B
ψ =0
q
A
x
Solution 6.22
∂φ
(a) u =
= 2x
∂x
In terms of the stream function,
u=
∂ψ
= 2x
∂y
Integrating with respect to y
 dψ =  2xdy
or
ψ = 2 xy + f1 ( x )
(1)
Similarly,
v=−
∂ψ ∂φ
=
= −2 y
∂x ∂y


So that dψ = 2 ydx
or
ψ = 2 xy + f2 ( y )
(2)
To satisfy both Eqs. (1) and (2)
ψ = 2xy + C
where C is an arbitrary constant. Since ψ = 0 along y = 0, C = 0 and
ψ = 2xy
(3)
(b) The discharge, q, passing through any surface connecting the two walls, such as AB
(see figure), is
q = ψ B −ψ A
From Eq. (3), ψ A = 0 and ψ B = 2 xi yi . It follows that
q = 2xi yi
Problem 6.23
A fluid with a density of 2000
kg
flows steadily between two flat plates as shown in the figm3
ure below. The bottom plate is fixed and the top one moves at a constant speed in the x direction. The velocity is V = 0.20 yˆi
m
where y is in meters. The acceleration of gravity is
s
m
g = −9.8ˆj 2 . The only nonzero shear stresses, τ yx = τ xy, are constant throughout the flow
s
N
with a value of 5 2 . The normal stress at the origin ( x = y = 0) is σ xx = −100 kPa . Use the
m
x and y components of the equations of motion
ρ gx +
ρ gy +
 ∂u
∂σ xx ∂τ yx ∂τ zx
∂u
∂u
∂u 
+
+
= ρ +u
+v
+w 
∂x
∂y
∂z
∂x
∂y
∂z 
 ∂t
∂τ xy
∂x
+
∂σ yy
∂y
+
∂τ zy
 ∂v
∂v
∂v
∂v 
= ρ +u
+v
+w 
∂z
∂x
∂y
∂z 
 ∂t
to determine the normal stress throughout the fluid. Assume that σ xx = σ yy.
Moving
plate
U
b
g
y
x
z
Fixed
plate
Solution 6.23
ρ gx +
ρ gy +
 ∂u
∂σ xx ∂τ yx ∂τ zx
∂u
∂u
∂u 
+
+
= ρ
+u
+v
+ w  (1)
∂x
∂y
∂z
∂x
∂y
∂z 
 ∂t
∂τ xy
∂x
+
∂σ yy
∂y
+
∂τ zy
 ∂v
∂v
∂v
∂v 
= ρ +u
+v
+ w  (2)
∂z
∂x
∂y
∂z 
 ∂t
For the given flow: u = 0.20 y
m
N
, v = 0, w = 0, τ xy = τ yx = 5 2 , τ zx = τ zy = 0 , g x = g z = 0,
s
m
m
and g y = −9.8 2
s
∂τ yx ∂τ zx
∂u
∂u
∂u ∂u
Thus,
= 0,
=
= 0, and v = 0 so that Eq. (1) becomes simply
=
= 0,
∂y
∂t
∂x ∂z
∂y
∂z
∂σ xx
= 0, or σ xx = f ( y ) + C , where C = constant and f ( y ) is an unknown function of y.
∂x
Similarly, since σ yy = σ xx, Eq. (1) becomes:
ρ gy +
∂σ xx
= 0 or
∂y
∂σ xx
kg 
m
kg
= − ρ g y = −2000 3  −9.8 2  = 19.6 ×103 2 2
∂y
m 
s 
m s
but
∂σ xx ∂f
=
∂y
∂y
so that
∂f
kg
= 19.6 ×103 2 2
∂y
m s
or
f = 19.6 ×103 y
kg ⋅ m
2 2
m s
+ const. = 19.6 ×103 y
N
m2
+ const.
Hence, since σ xx = f + C
σ xx = 19.6 y kPa + const. But σ xx = −100 kPa at y = x = 0 so that
σ xx = (19.6y −100) kPa, where y ∼ m.
Note: this is the standard hydrostatic pressure distribution.
Problem 6.24
In Section 6.3, we derived the differential equation(s) of linear momentum by considering
the motion of a fluid element. Derive the linear momentum equation(s) by considering a
small control volume, like we did for the continuity equation in Section 6.2.
Solution 6.24
Consider the differential control volume shown below (adapted from Fig. 6.5)
ρ vu +
ρ u2 –
∂ρ vu δ y
δ xδ z
∂v 2
∂ρ u2 δ x
δ yδ z
∂x 2
δy
ρ u2
y
δz
ρ u2 –
∂ρ u2 δ x
δ yδ z
∂x 2
δx
∂ρ vu δ y
δ xδ z
ρ vu +
∂y 2
x
z
The arrows show the x-momentum vectors at the left and right and top and bottom of
the control volume. The x-momentum vectors at the front and rear faces are omitted for
clarity.
∂M x
+  M x,out −  M x,in =  Fx where M x is the x-momentum;
∂t
M x = mu and M x is the flowrate of x-momentum; M x = mu. For the infinitesimal control
volume, M x = ρ (δ xδ yδ z )u and
For a control volume
 M x,out = [ mu ] right + [ mu ]top =  ρ u (δ yδ z ) u  right +  ρv (δ yδ z ) u  top =
 ρ u 2 (δ yδ z ) 
+ ρ vu(δ yδ z )] top

 right [
∂ρ u 2 δ x
= ρ u2 +
and
From the figure  ρ u 2 

 right
∂x 2
∂ρ u 2 δ x
and
Likewise  ρ u 2  = ρ u 2 −

 left
∂x 2
[ ρvu ]top = ρvu +
[ ρvu ] bottom = ρvu −
∂ρ vu δ y
∂y 2
∂ρ vu δ y
∂y 2
Then


∂M x
∂ρ u
∂ρ u 2 δ x 
∂ρ vu δ y 
δ xδ yδ z +  ρ u 2 +
δ xδ z
+  M x,out −  M x,in =
 δ yδ z +  ρvu +
∂t
∂t
∂x 2 
∂y 2 




∂ρ u 2 δ x 
∂ρvu δ y 
−  ρ u2 −
δ xδ z
 δ yδ z −  ρvu −
∂x 2 
∂y 2 


Cancelling and collecting terms
 ∂ρ u ∂ρ u 2 ∂ρ vu 
∂M x
+  M x,out −  M x,in = 
+
+
 δ xδ yδ z
∂t
∂x
∂x 
 ∂t
The obvious extension to three-dimensional flow is
 ∂ρ u ∂ρ u 2 ∂ρ vu ∂ρ wu 
∂M x
+  M x,out −  M x,in = 
+
+
+
 δ xδ yδ z
∂t
∂x
∂x
∂z 
 ∂t
The sum of the forces acting on the cube of fluid inside the control volume is identical to
the sum of forces when the cube is considered to be an individual particle and is given by
Eq. (6.50a) and shown in Fig. 6.11

∂σ
∂τ yx
∂τ

 Fx =  ρ gx + ∂xxx + ∂y + ∂zzx  δ xδ yδ z


Assembling the momentum equation and dividing by δ xδ yδ z gives the x-momentum equation for three-dimensional flow.
∂τ yx ∂τ zx
∂σ
∂ρ u ∂ρ u 2 ∂ρvu ∂ρ wu
+
+
+
= ρ gx + xx +
+
∂t
∂x
∂y
∂z
∂x
∂y
∂z
The y- and z-momentum equations are derived in the same way. They are
∂σ yy ∂τ xy ∂τ zy
∂ρ v ∂ρ uv ∂ρv 2 ∂ρ wv
+
+
+
= ρ gy +
+
+
∂t
∂x
∂y
∂z
∂y
∂x
∂z
∂τ yz ∂τ xz
∂σ
∂ρ w ∂ρ uw ∂ρvw ∂ρ w 2
+
+
+
= ρ g z + zz +
+
∂t
∂x
∂y
∂z
∂z
∂y
∂x
These equations are not identical to Eqs. (6.50). The momentum equations here are said to
be in conservative form. This form is most suitable for Computational Fluid Dynamics
(CFD). See Appendix A.
To illustrate the equivalence of the conservative form with Eqs. (6.50), consider the xmomentum equation. The difference is on the left (acceleration) side only. Using the Product Rule for Differentiation:
∂ρ u ∂ρ u 2 ∂ρvu ∂ρ wu
∂ρ ∂u
∂ρ u
∂u
∂ρ v ∂u
∂ρ w
∂u
+
+
+
=u
ρ +u
ρu + u
ρv + u
+ ρw
∂t
∂x
∂y
∂z
∂t ∂t
∂x
∂x
∂y
∂y
∂z
∂z
Grouping terms
 ∂ρ ∂ ρ u ∂ ρ v ∂ρ w 
 ∂u
∂ρ u ∂ρ u 2 ∂ρ vu ∂ρ wu
∂u
∂u
∂u 
+
+
+
=u
+
+
+
+ρ
+u
+v
+w 

∂t
∂x
∂y
∂z
∂x
∂y
∂z 
∂x
∂y
∂z 
 ∂t
 ∂t
According to the continuity equation [Eq. (6.27)]
 ∂ρ ∂ρ u ∂ρv ∂ρ w 
 ∂t + ∂x + ∂y + ∂z  = 0


 ∂u
∂ρ u ∂ρ u 2 ∂ρ vu ∂ρ wu
∂u
∂u
∂u 
+
+
+
= ρ
+u
+v
+ w  . The same
∂t
∂x
∂y
∂z
∂x
∂y
∂z 
 ∂t
manipulation can be applied to the y- and z-momentum equations so the three momentum
equations can be written
So
∂τ yx ∂τ zx
 ∂u
∂σ
∂u
∂u
∂u 
+u
+ v + w  = ρ gx + xx +
+
∂x
∂y
∂z 
∂x
∂y
∂z
 ∂t
ρ
∂σ yy ∂τ xy ∂τ zy
 ∂v
∂v
∂v
∂v 
+u
+ v + w  = ρ gy +
+
+
∂x
∂y
∂z 
∂y
∂x
∂z
 ∂t
ρ
 ∂w
∂σ zz ∂τ yz ∂τ xz
∂w
∂w
∂w 
g
ρ
+u
+v
+w
=
+
+
+
z

∂x
∂y
∂z 
∂z
∂y
∂x
 ∂t
ρ
which are equivalent to Eqs. (6.50). Options for the stress terms on the right side are the
same as those discussed in Chapter 6 for laminar flow and Appendix A for turbulent flow.
Comment The first equations derived here are said to be in conservative form because they
contain both conservation of mass and “conservation” of momentum.
Problem 6.25
By considering the rotational equilibrium of a fluid mass element, show that τ xy = τ yx .
Solution 6.25
The sketch shows the stresses in the x − y plane acting on a fluid particle.
σyy
τ yx
τ xy
δy
σxx
τ xy
σxx
δx
τ yx
σyy
Apply Newton’s second law in rotational form
 M z = I zωz
where M z is the moment about the z axis (perpendicular to the paper, passing through the
center of the element), I z the moment of inertia, and ω z the angular acceleration.
For the element (Note the normal stresses produce no moment):
(τ xyδ yδ z )δ x − (τ yxδ xδ z )δ y = ρ (δ xδ yδ z )([ kδ x ] )ω
2
where k is a constant and kδ x is the radius of gyration.
Dividing by δ xδ yδ z
τ yx − τ xy = k 2 (δ x )2 ω 2
k is finite (typically k <1 ) and ω is finite so as the fluid element shrinks to a point (δ x → 0),
the right side vanishes and
τ yx − τ xy = 0
or
τ yx = τ xy
Problem 6.26
The stream function for a given two-dimensional flow field is
ψ = 5 x 2 y − 5 y3
Determine the corresponding velocity potential.
Solution 6.26
u=
∂ψ ∂φ
=
= 5x 2 − 5 y 2
∂y ∂x
(1)
Integrate with respect to x to obtain
 dφ = − 5x − 5 y dx
2
2
or
5
3
φ = x3 − 5xy2 + f1 ( y )
(2)
Similarly,
v=−
∂ψ ∂φ
=
= −10xy
∂x ∂y
(3)
and
 dφ = −10xydy
or
φ = −5xy2 + f2 ( x )
To satisfy both Eqs. (2) and (4)
5
3
φ = ( )x3 − 5xy2 + C
where C is an arbitrary constant.
(4)
Problem 6.27
A certain flow field is described by the stream function
ψ = Aθ + Br sin θ
where A and B are positive constants. Determine the corresponding velocity potential and
locate any stagnation points in this flow field.
Solution 6.27
vr =
1 ∂ψ ∂φ A
=
= + B cos θ
r ∂θ ∂r r
(1)
Integrate with respect to r to obtain
A

 dφ =   r + B cos θ  dr
or
φ = A ln r + Br cos θ + f1 (θ )
(2)
Similarly,
vθ = −
∂ψ 1 ∂φ
=
= − B sin θ
∂r r ∂θ
(3)
and
 dφ = − Br sinθ dθ
or
φ = Br cos θ + f2 ( r )
(4)
To satisfy both Eqs. (2) and (4)
φ = A ln r + Br cos θ + C
where C is an arbitrary constant.
Stagnation points occur where vr = 0 and vθ = 0 .
From Eq. (3), vθ = 0 at θ = 0 and θ = π . From Eq. (1), with θ = 0
vr =
A
+B
r
A
So that vr = 0 for r = − . However, since A and B are both positive constants, this result
B
indicates a negative value for r which is not defined.
At θ = π
A
A
vr = + B cos π = − B
r
r
A
So that vr = 0 for r = . Thus, a stagnation point occurs at
B
A
θ = π and r =
B
Problem 6.28
Integrate Bernoulli’s equation for compressible flow,
dp V 2
 ρ + 2 + gz = constant ,
for an ideal gas undergoing an isothermal (constant temperature) process along a streamline.
Solution 6.28
The equation along a streamline is
dp V 2
 ρ + 2 + gz = C
From the ideal gas low
ρ=
p
RT
so
1
ρ
=
RT
p
Substituting the idea gas law into the first term of Bernoulli’s equation gives
dp
dp
dp
 ρ =  RT p = RT  p = RT ln p
for a flow at constant temperature. The result is
V2
RT ln p +
+ gz = C along a streamline
2
Problem 6.29
Integrate Bernoulli’s equation for compressible flow,
dp V 2
 ρ + 2 + gz = constant ,
for a fluid whose pressure p and density ρ obey the polytropic expression p = C ρ n along a
streamline, where C and n are constants.
Solution 6.29
The equation is
dp V 2
 ρ + 2 + gz = C
From the given pressure–density relationship
p = C ρ n , dp = nC ρ n −1d ρ ,
And the first term of Bernoulli’s equation becomes
dp
ρ
=  nC ρ n−2 d ρ = nC
But C =
p
dp
ρ
ρ
so 
n
=
ρ n−1
n −1
n p n−1
n p
ρ
=
n −1 ρ n
n −1 ρ
The integrated equation is
n p V2
+
+ gz = C
n −1 ρ
2
.
Problem 6.30
The velocity potential for a certain inviscid flow field is
φ = −(3x 2 y − y3 )
ft 2
when x and y are in feet. Determine the pressure difference (in
s
psi) between the points (1, 2) and (4, 4), where the coordinates are in feet, if the fluid is water and elevation changes are negligible.
where φ has the units of
Solution 6.30
Since the flow field is described by a velocity potential the flow is irrotational and the Bernoulli equation can be applied between any two points. Thus,
p1 V12 p2 V22
+
=
+
γ 2g γ 2g
∂φ
= −6xy
Also, u =
∂x
(1)
v=
∂φ
= −3x2 + 3 y2
∂y
At x = 1ft , y = 2 ft
u1 = −6(1)(2) = −12
ft
s
v1 = −3(1)2 + 3(2)2 = 9
ft
s
2
2
ft   ft 

 ft 
So that V12 = u12 + v12 =  −12  +  9  = 225  
s  s

s
at x = 4 ft , y = 4 ft
u2 = −6(4)(4) = −96
ft
s
v2 = −3(4)2 + 3(4)2 = 0
ft 

So that v2 2 =  −96 
s

2
Thus, from Eq. (1)
p1 − p2 =
1γ
[V22 −V12 ]
2g
1b 

62.4 3  
2

1
ft 

 ft  
ft

=
 −96  − 225   2 
ft  
2 
s
 s  
 32.2 2  
s 

= 8710
1b   ft 2 

=
8710
 = 60.5 psi


ft 2 
ft 2   144 in2 
1b
2
Problem 6.31
The stream function for a two-dimensional, nonviscous, incompressible flow field is given
by the expression
ψ = −2( x − y )
where the stream function has the units of ft2/s with x and y in feet. (a) Is the continuity
equation satisfied? (b) Is the flow field irrotational? If, so, determine the corresponding
velocity potential. (c) Determine the pressure gradient in the horizontal x direction at the
point x = 2 ft,y = 2 ft .
Solution 6.31
(a) To satisfy the continuity equation,
∂u ∂v
+
=0
∂x ∂y
For the stream function given,
u=
∂ψ
ft
=2
∂y
s
v=−
∂ψ
ft
=2
∂x
s
So that
∂v
∂u
=0
= 0 and
∂y
∂x
so the continuity equation is satisfied. Yes.
(Note: when a flow field is defined by a stream function the continuity equation is always
identically satisfied.)
(b) Since
1  ∂v
∂u 
ωz =  − 
2  ∂x ∂y 
And
∂u
=0
∂y
∂v
=0
∂x
It follows that ω z = 0 and the flow field is irrotational. Yes.
Thus, u =
∂φ
=2
∂x
v=
∂φ
=2
∂y
And integration yields
φ = 2(x + y) + C
where C is an arbitrary constant.
(c) With the x-axis horizontal, g x = 0 , and
−
 ∂u
∂p
∂u 
= ρ u
+v 
∂x
∂y 
 ∂x
And at x = 2 ft,y = 2 ft
∂p
ft
 ft

= − ρ  2 (0) + 2 (0)  = 0
s
∂x
 s

Problem 6.32
The velocity potential for a certain inviscid, incompressible flow field is given by the equation
2
φ = 2 x 2 y −   y3
3
where φ has the units of m2/s when x and y are in meters. Determine the pressure at the
point x = 2 m,y = 2 m if the pressure at x = 1m,y = 1 m is 200 kPa. Elevation changes can be
neglected, and the fluid is water.
Solution 6.32
Since the flow is irrotational,
p1 V12 p2 V2 2
+
=
+
γ 2g γ
2g
(1)
With V 2 = u 2 + v 2. For the velocity potential given,
u=
∂φ
= 4 xy
∂x
v=
∂φ
= 2 x 2 − 2 y2
∂y
At point 1, let x = 1m and y = 1 m so that
u1 = 4(1)(1) = 4
m
, v1 = 2(1)2 − 2(1)2 = 0
s
And
2
m2
 m
V12 =  4  = 16 2
 s 
s
At point 2, x = 2 m and y = 2 m so that
u2 = 4(2)(2) = 16
m
, v2 = 2(2)2 − 2(2)2 = 0
s
And
2
m2
 m
V2 2 =  16  = 256 2
s 

s
Thus, from Eq. (1)
p2 = p1 +
γ
2g
(V12 −V22 )
N 

9.80 ×103 3 

2
2
N
m   16 m − 256 m 
= 200 ×103 2 + 

2
2 

m 

m
s
s

2  9.81 2 
s 

= 80.1 kPa
Problem 6.33
Consider the two-dimensional air flow around the corner, as shown in the figure below. The
v
x
y
x- and y-direction velocities are
u = 0 sinh   cosh  
L
L
L
v
x
y
and
v = − 0 cosh   sinh   ,
L
L
L
respectively. Assume constant density, steady flow, negligible gravity, and inviscid flow.
Find p ( x, y ).
y
v (x, L)
L
0
u (L, y )
x
L
0
Solution 6.33
Euler’s equation in Cartesian coordinates for two-dimensional flow are
u
∂u
∂u
1 ∂p
+v = −
ρ ∂x
∂x
∂y
and
u
∂v
∂v
1 ∂p
+v = −
ρ ∂y
∂x
∂y
Now
∂u V0
x
y
= 2 cosh   cosh   ,
∂x L
L
L
∂u V0
x
y
= 2 sinh   sinh   ,
∂y L
L
L
V
∂v
x
y
= − 0 sinh   sinh   ,
L
∂x
L
L
V
∂v
x
y
= − 02 cosh   cosh  
∂y
L
L
L
Substituting into Euler’s equation gives.
V0
x
 y   V0
x
 y 
 L sinh  L  cosh  L    2 cosh  L  cosh  L  
 
   L
 
 

 V
1 ∂p
x
 y   V
x
 y 
+  − 0 cosh   sinh     02 sinh   sinh    = −
ρ ∂x
L
 L   L
L
 L 
 L
and
V0
x
 y    V0
x
 y 
 L sinh  L  cosh  L    − 2 sinh  L  sinh  L  
 
   L
 
 

 V
1 ∂p
x
 y   V
x
 y 
+  − 0 cosh   sinh     − 02 cosh   cosh    = −
ρ ∂y
L
 L   L
L
 L 
 L
Simplifying the equation for
∂p
gives
∂x
V0 2 
1 ∂p
x
x
y
x
x
 y 
,
sinh   cosh   cosh 2   − cosh   sinh   sinh 2    = −
3 
ρ ∂x
L 
L
L
L
L
L
 L 
1 ∂p
x
 x 
2 y
2  y 
,
sinh
cosh
cosh
−
sinh
=
−










ρ ∂x
L3
L
 L 
L
 L 
V0 2
or
V02
1 ∂p
 2x 
sinh   = −
(1)
ρ ∂x
2L
 L 
3
Simplifying the equation for
∂p
gives
∂y
V0 2 
1 ∂p
y
y
y
 y 
2 x
2 x
,
−
sinh
cosh
sinh
+
cosh
sinh
cosh
=
−














ρ ∂y
L3 
L
L
L
L
L
 L 
1 ∂p
y
 y 
2x
2  x 
,
sinh
cosh
cosh
−
sinh
=
−










ρ ∂y
L3
L
 L 
L
 L 
V0 2
or
V02
1 ∂p
 2y 
sinh   = −
(2)
ρ ∂y
2L
 L
3
Integrating Eqs. (1) and (2) gives
 dp = −
ρV0 2
 2x 
 sinh  L  dx + f1 ( y )
2 L3
or
p=−
ρV0 2
 2x 
cosh 
 + f1 ( y )
 L 
4L
2
(3)
and
 dp = −
ρV0 2
 2y 
 sinh  L  dx + f2 ( x )
2L
3
or
p=−
ρV02
 2y 
cosh   + f2 ( x )
4
 L 
Equations (3) and (4) give
p=−
ρV02 
4
 2x 
 2 y 
cosh  L  cosh  L   + const.
 
 

(4)
Comment. The term V 0 was not defined in the problem statement, but it could not be a
velocity. It is a volume flowrate per unit length perpendicular to the page. Hence
L3 L2
[V0 ] =
=
LT T
and
2
V0   L   1  L
=

 L   T   L  = T
    
Problem 6.34
The streamlines for an incompressible, inviscid, two-dimensional flow field are all concentric circles, and the velocity varies directly with the distance from the common center of the
streamlines; that is
vθ = Kr
where K is a constant. (a) For this rotational flow, determine, if possible, the stream
function. (b) Can the pressure difference between the origin and any other point be
determined from the Bernoulli equation? Explain.
Solution 6.34
(a) vθ = −
∂ψ
= Kr
∂r
(1)
Integrate Eq. (1) with respect to r to obtain
 ∂ψ = − Krdr
or
ψ =−
Kr 2
+ f1(θ )
2
Since
vr =
1 ∂ψ
=0
r ∂θ
It follows that ψ is not a function of θ and therefore
ψ =−
Kr 2
+C
2
where C is an arbitrary constant.
(b) The flow is rotational and therefore the Bernoulli equation cannot be applied between the origin and any point, since these points are not on the same streamline.
No.
(Refer to discussion associated with derivation of the following equation
p V2
+
+ gz = constant along a streamline)
ρ 2
Problem 6.35
The velocity potential
φ = − k ( x 2 − y 2 )(k = constant)
may be used to represent the flow against an infinite plane boundary, as illustrated in the
figure below. For flow in the vicinity of a stagnation point, it is frequently assumed that the
pressure gradient along the surface is of the form
∂p
= Ax
∂x
where A is a constant. Use the given velocity potential to show that this is true.
y
x
Solution 6.35
For the velocity potential given
∂φ
= −2 kx
∂x
∂φ
v=
= 2ky
∂y
u=
(1)
(2)
and the stagnation point occurs at the origin. For this steady, two-dimensional flow
−
 ∂u
∂p
∂u 
= ρ u
+v 
∂x
∂y 
 ∂x
and along the surface (y = 0) v = 0 so that
∂p
∂u
= ρu
∂x
∂x
(3)
From Eq. (1) u = −2kx
Therefore
∂u
= −2k
∂x
And Eq. (3) becomes
∂p
= ρ ( −2 kx )( −2 k ) = 4 k 2 x
∂x
or
∂p
= Ax ,
∂x
where A = 4 k 2
Problem 6.36
Water is flowing between wedge-shaped walls into a small opening as shown in the figure
m2
for this flow is φ = −2 ln r with r in meters. Debelow. The velocity potential with units
s
termine the pressure differential between points A and B.
_π_
6
r
A θ
0.5 m
B
1.0 m
Solution 6.36
pA VA2 pB VB 2
+
=
+
γ
2g
γ
2g
(1)
Along the horizontal surface, vθ = 0 , and vr =
So that V = vr = −
Thus, VA = −
2
r
2
m
= −4
0.5
s
VB = −
2
4m
=−
1.5
3 s
And from Eq. (1)
pA − pB =
γ
2g
[VB 2 −VA2 ]
N
2
2

m3  − 4 m  −  −4 m  
=

m   3 s  
s  

2  9.81 2  
s 

9.80 ×103
= −7.10 kPa
∂φ
2
=−
r
∂r
Problem 6.37
The velocity potential for a given two-dimensional flow field is
5
φ =   x3 − 5 xy 2
3
Show that the continuity equation is satisfied and determine the corresponding stream
function.
Solution 6.37
To satisfy the continuity equation,
∂u ∂v
+
=0
∂x ∂y
For the given velocity potential,
∂φ
5
= (3)   x 2 − 5 y 2 = 5 x 2 − 5 y 2
∂x
3
u=
and
∂u
= 10 x
∂x
Similarly,
v=
∂φ
= −10xy
∂y
and
∂v
= −10x
∂y
Thus,
∂u ∂v
+ = 10x − 10x = 0
∂x ∂y
and the continuity equation is satisfied.
Since u =
∂ψ
= 5x 2 − 5 y 2
∂y
and integrating with respect to y gives
 dψ =  (5x − 5 y )dy
2
2
or

ψ = 5  x 2 y −

y3 
 + f1( x )
3 
(1)
Similarly, v = −
∂ψ
= 10xy
∂x
And integrating with respect to x gives
ψ = 5 x 2 y + f2 ( y )
To satisfy both Eqs. (1) and (2),
5
3
ψ = 5x 2 y − y3 + C with C a constant.
(2)
Problem 6.38
As illustrated in the figure below, a tornado can be approximated by a free vortex of
strength for r > Rc , where Rc is the radius of the core. Velocity measurements at points A
ft
ft
and B indicate that VA = 125 and VB = 60 . Determine the distance from point A to the
s
s
center of the tornado. Why can the free vortex model not be used to approximate the tornado throughout the flow field (r ≥ 0 ) ?
y
r
Rc
A
B
x
100 ft
Solution 6.38
For a free vortex
vθ =
k
r
Thus, at rA , vθ = 125
and at rB , vθ = 60
ft
, so that K = 125 rA
s
ft
, so that K = 60 rB.
s
Therefore,
125 rA = 60 rB
and since
rB − rA = 100 ft
it follows that
125 rA = 60 (100 + rA )
or
rA = 92.3ft
The free vortex cannot be used to approximate a tornado throughout the flow field since at
r = 0 the velocity becomes infinite.
Problem 6.39
The motion of a liquid in an open tank is that of a combined vortex consisting of a forced
vortex for 0 ≤ r ≤ 2 ft and a free vortex for r > 2 ft. The velocity profile and the corresponding shape of the free surface are shown in the figure below. The free surface at the center of
the tank is a depth h below the free surface at r = ∞. Determine the value of h. Note that
h = h forced + h free , where h forced and h free are the corresponding depths for the forced vortex
and the free vortex, respectively.
10
vθ , ft/s
0
r, ft
2
z
2
r, ft
h
Solution 6.39
For forced vortex
ω 2r2
z=
+C
2g
and with z = 0 at r = 0, it follows that C = 0.
ft
Also, vθ = rω and since vθ = 10 at r = 2ft
s
ft
10
rad
ω = s =5
2 ft
s
Thus, at r = 2 ft
2
 rad 
5
(2 ft)2
2 2


ω r
s 
z=
=
= 1.55 ft
ft 
2g

2  32.2 2 
s 

z
2
h
zs
r, ft
For free vortex
zs =
Γ2
8π 2 r 2 g
where Γ = 2π rvθ
So that
2
2
ft 
4π (2 ft)  10 
2
(2π rvθ )
 s  = 1.55 ft
=
zs =
2 2
ft 

8π r g
8π 2 (2 ft)2  32.2 2 
s 

2
Thus,
h = h forced + h free = 1.55ft + 1.55ft = 3.10 ft
Problem 6.40
Water flows through a two-dimensional diffuser having a 20 expansion angle as shown in
the figure below. Assume that the flow in the diffuser can be treated as a radial flow
m
emanating from a source at the origin O . (a) If the velocity at the entrance is 20 ,
s
determine an expression for the pressure gradient along the diffuser walls. (b) What is the
pressure rise between the entrance and exit?
Diffuser wall
2m
r
20°
O
Flow
Entrance
7m
Exit
Solution 6.40
(a) For radial flow
vr =
m
2π r
For r = 2m, vr = 20
m
so that
s
m
 m
m = 2π rvr = 2π (2 m)  20  = 80π
s 
s

2
From the Bernoulli equation
1
p + ρvr 2 = constant
2
So that
∂v
∂p
= − ρvr r
∂r
∂r
Since vr =
(1)
∂v
m
m
then r = −
2π r
∂r
2π r 2
and Eq. (1) can be written as
∂p
m  ρ m2
 m 
= −ρ 
 −
=
∂r
 2π r   2π r 2  4π 2 r3
Thus, for ρ = 999
∂p
=
∂r
kg
m
, m = 80π
3
kg  
m2 

999
80
π



s 
m3  

4π 2 r3
m2
s
2
=
1.60 ×106 N
r3
m
=
3
1.60 × 103 kPa
m
r3
(b) Since (vr )entrance = 20
m
and
s
m2
m
s = 5.71 m
(vr )exit =
=
2π r 2π (7 m )
s
80π
Then from the Bernoulli equation
pexit − pentrance =
1
ρ [(vr )2 entrance − (vr )2exit ]
2
2
2
1
kg   m  
m 
=  999 3   20  −  5.71   = 184 kPa
2
s  
s  
m  
Problem 6.41
When water discharges from a tank through an opening in its bottom, a vortex may form
with a curved surface profile, as shown in the figure below. Assume that the velocity
distribution in the vortex is the same as that for a free vortex. At the same time, the water is
being discharged from the tank at point A, it is desired to discharge a small quantity of
water through the pipe B. As the discharge through A is increased, the strength of the
vortex, as indicated by its circulation, is increased. Determine the maximum strength that
the vortex can have in order that no air is sucked in at B. Express your answer in terms of
the circulation. Assume that the fluid level in the tank at a large distance from the opening
at A remains constant and viscous effects are negligible.
B
1 ft
2 ft
A
Solution 6.41
For free vortex (from Example 6.6):
zs = −
Γ2
8π 2 r 2 g
Air will be sucked into pipe when z s = −1ft for r = 2 ft.
ft 
ft 


Thus, Γ2 = −8π 2r 2 gzs = −8π 2 (2 ft)2  32.2 2  ( −1ft)=32π 2  32.2 2 
s 
s 


or
Γ = 100.8
ft 2
s
Problem 6.42
Water flows over a flat surface at 4 ft/s, as shown in the figure below. A pump draws off
water through a narrow slit at a volume rate of 0.1 ft3/s per foot length of the slit. Assume
that the fluid is incompressible and inviscid and can be represented by the combination of a
uniform flow and a sink. Locate the stagnation point on the wall (point A) and determine
the equation for the stagnation streamline. How far above the surface, H, must the fluid be
so that it does not get sucked into the slit?
4 ft/s
H
A
0.1 ft3/s
(per foot of length of slit)
Solution 6.42
V
4 ft/s
H
θ
A
rs
3
0.1 ft /s
(per foot of length of slit)
ψ = ψ uniform +ψ sink = Ur sin θ −
flow
m
θ
2π
(1)
Thus,
vr =
1 ∂ψ
m
= U cos θ −
r ∂θ
2π r
(2)
and
vθ = −
∂ψ
= −U sin θ
∂r
Along the wall vθ = 0 , and the stagnation point occurs where vr = 0 , so that from Eq. (2)
0 = U cos(0 ) −
and therefore
rs =
m
2πU
m
2π rs
ft 2
ft2
ft
For U = 4 and m = 0.2
(note that a source strength of 0.2
must be used to obtain
s
s
s
ft 2
0.1
through slit which is only one half of a “full” sink). Thus,
s
ft 2
s = 0.00796 ft
rs =
 ft 
2π  4 
 s
0.2
and the stagnation point is on the wall 0.00796 ft to the right of slit.
The value of ψ at the stagnation point (r = 0.00796 ft, θ = 0 ) is zero [Eq. (1)] so that the
equation of the stagnation streamline is:
0 = Ur sin θ −
m
θ
2π
or
r sinθ =
m
θ
2π v
Since y = r sin θ , the equation of the stagnation streamline can be written as
y=
m
θ
2πU
Fluid above the stagnation streamline will not be sucked into slit. The maximum distance,
H, for the stagnation streamline occurs as θ → π so that
ft 2
mπ
s = 0.0250 ft
=
H=
2πU
 ft 
24 
 s
0.2
(Note: All the fluid below the stagnation streamline must pass through the slit. Thus, from
conservation of mass
HU = flow into slit
or
ft 2
s = 0.0250 ft
H=
ft
4
s
0.1
which checks with the answer above.)
Problem 6.43
Two sources, one of strength m and the other with strength 3m, are located on the x axis as
shown in the figure below. Determine the location of the stagnation point in the flow
produced by these sources.
y
2ℓ
3ℓ
x
+m
+3m
Solution 6.43
Since the flow from each source is in the radial direction, it is only along the x-axis that the
two radial components can cancel and create a stagnation point.
y
(1)
2ℓ
3ℓ
r1stag
r2stag
vr1
Stagnation
point
+m
(2)
vr 2
x
+3m
For source (1)
vr1 =
m
2π r1
and for source (2)
vr2 =
3m
2π r2
The stagnation point occurs where vr1 = vr 2 so that
m
3m
=
2π r1stag 2π r2 stag
and
r2 stag
r1stag
=3
Also,
r1stag + r2stag = 2l + 3l = 5l
So that
r1stag + 3r1stag = 5l
5
r1stag = l
4
5
Thus, xstag = −  2l − l  = −0.75l

4 
Problem 6.44
 Γ 
 m 
The velocity potential for a spiral vortex flow is given by φ = 
θ − 
 ln r , where Γ
 2π 
 2π 
and m are constants. Show that the angle, α , between the velocity vector and the radial direction is constant throughout the flow field (see the figure below).
y
V
α
r
θ
x
Solution 6.44
For the velocity potential given,
vr =
m
∂φ
=−
2π r
∂r
vθ =
1 ∂φ
Γ
=
r ∂θ 2π r
y
êθ
V
α
r
eˆ r
θ
x
Since
V ⋅ eˆ r =| V | cos α
and
V = vr eˆ r + vθ eˆ θ
Then
cos α =
Veˆ r
vr
=
2
|V|
vr 2 + vθ
1
1
=
=
2
2
 vθ 
 Γ 
1+  


2π r 
 vr 
1+ 
2
 m 
−

 2π r 
=
1
Γ
1+  
m
2
Thus, for a given Γ and m the angle α is a constant .
Problem 6.45
For a free vortex, determine an expression for the pressure gradient (a) along a streamline,
and (b) normal to a streamline. Assume that the streamline is in a horizontal plane, and
express your answer in terms of the circulation.
Solution 6.45
For a free vortex [Eq. (6.91)]:
Γ
ψ = − ln r
2π
1 ∂φ
∂φ
Γ
So that vr =
vθ = −
=0
=
∂r 2π r
r ∂θ
Since the free vortex represents an irrotational flow field, the Bernoulli equation
p V2
+
+ z = constant
γ 2g
(1)
is valid between any two points.
(a) Along a streamline (r = constant), vθ is constant and vr = 0 so that from Eq. (1) with
z constant the pressure is constant, i.e.,
∂p
=0
∂θ
(b) Normal to the streamline with vr = 0 and z = constant
v 2
+ θ + z = constant
γ 2g
p
So that
∂v
∂p
γ ∂(vθ 2 )
=−
= − ρvθ θ
∂r
2 g ∂r
∂r
Γ 
 Γ 
= −ρ 
 −

 2π r  2π r 2 
ρ Γ2
= 2 3
4π r
Problem 6.46
Consider a category five hurricane that has a maximum wind speed of 160 mph at the eye
wall, 10 mi from the center of the hurricane. If the flow in the hurricane outside of the hurricane’s eye is approximated as a free vortex, determine the wind speeds at locations 20 mi,
30 mi, and 40 mi from the center of the storm.
Solution 6.46
For free vortex
k
vθ =
r
Vθ
rB
rA
Eye wall
Thus, at the eye wall
k
160 mph =
10 mi
So that
k = (160 mph)(10 mi)
and
vθ =
(160 mph)(10 mi)
rB
For,
(160 mph)(10 mi)
= 80.0 mph
20 mi
(160 mph)(10 mi)
= 53.3 mph
rB = 30 mi vθ =
30 mi
(160 mph)(10 mi)
= 40.0 mph
rB = 40 mi vθ =
40 mi
rB = 20 mi vθ =
Problem 6.47
Consider the flow of a liquid of viscosity μ and density ρ down an inclined plate making an
angle θ with the horizontal. The film thickness is t and is constant. The fluid velocity parallel to the plate is given by
ρt2 g cos θ   y  
Vx =
1 −    ,
2μ
  t  
2
where y is the coordinate normal to the plate. Calculate
neither satisfies Laplace’s equation. Why not?
and
Solution 6.47
Since
2
ρt2 g cos θ   y  
∂φ
= Vx, then φ =
1 −    x + f ( y )
2μ
∂x
  t  
Since
∂ψ
ρt2 g cos θ 
y3 
= Vx, then ψ =
y − 2  + g ( x )
∂y
2μ
3t 

Since the density is constant, the continuity equation gives
∂Vy
∂Vx ∂Vy
∂V
+
= 0,
= − x = 0,
∂x
∂y
∂y
∂x
or
Vy = constant = C
From the boundary condition, Vy = 0 at y = 0
Vy = 0. Then
∂φ
∂ψ
= Vy = 0 and
= −Vy = 0
∂y
∂x
Then
ρt 2 g cos θ   y  
φ=
1 −    x + C1
2μ
  t  
2
And
ψ=
ρt 2 g cos θ 
y3 
 y − 2  + C2
2μ
3t 

where C 1 and C 2 are constant.
for this flow and show that
Then
∂ 2φ
∂x2
∂ 2φ
∂x
2
ρ g cos θ
ρ t 2 g cos θ  2 
,
=
− 2 =−

2
μ
2μ
∂y
 t 
∂ 2φ
= 0,
+
∂ 2φ
∂y
2
=−
ρ g cos θ
≠ 0,
μ
and
φ does not satisfy Laplace’s equation for this flow.
Next,
∂ 2ψ
∂x 2
∂ 2ψ
∂x
2
ρ g cos θ y
ρ t 2 g cos θ  2 y 
,
=
− 2 = −

2
μ
2μ
∂y
 t 
∂ 2ψ
= 0,
+
∂ 2ψ
∂y
2
=−
ρ g cos θ y
≠0
μ
and
ψ does not satisfy Laplace’s equation for this flow.
The reason that neither φ nor ψ satisfy Laplace’s equation is that the flow is not irrotational. It is a viscous flow and has rotation.
Problem 6.48
A horizontal oil-hearing stratum is 3 m high and, after hydraulic fracturing (“fracking”),
m3
. The flow moves radially inward and is collected
provides a volume flowrate of Q = 1000
hr
by a vertical porous pipe having an outer radius of 1.0 m . The Laplace equation for the potential flow model of the oil is
1  ∂ 2φ  ∂ 2φ 1  ∂ 2φ 
+ 

+
=0.
r  ∂r 2  ∂r 2 r 2  ∂θ 2 
Find φ .
Solution 6.48
Start with
1  ∂ 2φ  ∂ 2φ 1  ∂ 2φ 
+ 

+
=0
r  ∂r 2  ∂r 2 r 2  ∂θ 2 
or
 ∂ 2φ  ∂ 2φ 1  ∂ 2φ 
 2  + r 2 +  2  = 0
∂r
r  ∂θ 
 ∂r 
Use separation of variables with φ
θ =
θ
Substituting gives
R (r )
d 2P (θ )
dθ 2
+ r2 P ( θ )
d 2R ( r )
dr 2
+ rP ( θ )
d 2R ( r )
=0
dr
And
1 d 2P (θ )
1  2 d 2R ( r )
dR ( r ) 
2
r
+
r
=
−

 = −λ
2
dr
P ( θ ) dθ 2
R ( r ) 
dr

Then
d 2P (θ )
+ λ 2 P ( θ ) = 0, r 2
d 2R ( r )
+r
dR ( r )
− λ 2 R ( r ) = 0,
dr
dr
dθ
P ( θ ) = A cos( λθ ) + B sin( λθ ), and R ( r ) = Cr λ + Dr − λ
2
2
One boundary condition is Vr → 0 as r → ∞ and gives
∂φ
dR ( r )
dR ( r )
, lim Vr = lim P ( θ )
= Vr = P ( θ )
Vr →0
r →∞
dr
dr
∂r
dR ( r )
as r → ∞ . Therefore, C = 0 and
And 0 =
dr
R ( r ) = Dr − λ . A second boundary condition is φ ( r, c ) = φ ( r,2π ) and gives
P (0 )R ( r ) = P (2π )R ( r ) , P(0) = P(2π ),
A cos(0) + B sin(0) = A cos(2πλ ) + B sin(2πλ ), and
A = A cos(2πλ ) + B sin(2πλ ) with λ = 0,1,2,3... = n
Then
∞
φ ( r,θ ) =  An r − k cos( nθ ) + Bnr − k sin( nθ ),
n =0
Vr =
∞
∂φ
= − nr −( n+1) [ An cos( nθ ) + Bn sin( nθ )],
∂r
n =1
And V r at r = 1.0 m is found from
Q = (3 m)2π (1.0 m)Vr = 1000
m3
500 m
or Vr =
hr
3π hr
Which is negative because flow is inward.
Then −
∞
500
= − n[ An cos( nθ ) + Bn sin( nθ )]
3π
n =1
(1)
Use the orthogonality property of the trig functions where
0 m ≠ n
π
−π sin( mx )sin( nx )dx = π m ≠ n
π
−π sin( mx )cos( mx )dx =0 for all m, n and
0 m ≠ n
mx
nx
dx
cos(
)
cos(
)
=

−π
π m ≠ n
π
Multiplying Eq. (1) by cos( mθ )dθ and integrating gives
π ∞
π ∞
500
cos(
)
cos(
)cos(
)
m
θ
d
θ
=
+
nA
n
θ
m
θ
d
θ
+
−π 3π
−π  n
−π  nBn sin( nθ ) cos( mθ )dθ
n =1
n =1
π
Note that n = m
where 1 ≤ m ≤ ∞ in discrete intervals so
π 500
π
π
2
−π 3π cos( mθ )dθ = + −π mAm cos ( mθ )dθ + −π mBm sin( mθ ) cos( mθ )dθ
π
 500

and 
sin( mθ )  = + mAmπ + 0 .
 3π m
 −π
Then
500
[sin( mπ ) − sin( − mπ )] = mAmπ or 0 = Am.
3π m
Multiplying Eq. (1) by sin( mθ )dθ and integrating gives
π ∞
500
sin(
)
m
θ
d
θ
=
+
−π 3π
−π  nBn sin( nθ )sin( mθ )dθ
n =1
π
Again note that n = m where 1 ≤ m ≤ ∞ in discrete intervals so
π
500
π
−π 3π sin( mθ )dθ = + mBn −π sin ( mθ )dθ , and
−
500
cos( mθ )]π−π = + mBmπ
3π m
2
500
[cos( mπ ) − cos( mπ )] = − mBmπ , and 0 = Bm
3π m
Now let us check the case for λ = 0
d 2P (θ )
ds 2
= 0, r 2
d 2R ( r )
dr 2
+r
dR ( r )
=0
dr
P ( θ ) = C θ + D , and R = A + Blnr.
φ ( r , θ ) = P ( θ ) R ( r ) , Φ ( r , θ ) = Φ ( r , 2π )
P (0 )R ( r ) = P (2π )R ( r ) , P (0) = P (2π ),
D = 2πC + D, C = 0 so Φ ( r,θ ) = A + B ln r .
Now
Vr =
500 B
∂φ dφ B
=
at r = 1.0
=
= ,=−
3π 1.0
∂r dr r
500 B
500
=
so B =
.
3π 1.0
−3π
500
Then ρ ( r ) = −
ln R + A. (R=r)
3π
And −
Since all boundary conditions have been satisfied,
φ (r ) = −
500
ln R
3π
Problem 6.49
Show that the circulation of a free vortex for any closed path that does not enclose the
origin is zero.
Solution 6.49
Γ
θ
2π
The velocity components, in polar coordinates, for a free vortex are
The velocity potential for a free vortex is given by φ =
Vr =
∂φ ∂  Γ 
1 ∂φ 1 ∂  Γ 
Γ
= 
θ  = 0 and Vθ =
θ=
=

∂r ∂r  2π 
r ∂θ r ∂θ  2π  2π r
iy
D
A
C
B
x
A closed curve C, not enclosing the origin, is defined as shown in the sketch, in order to
more easily evaluate the integrals in the circulation expression.The circulation about the
closed curve C is given by
B
C
D
A
A
B
C
D
Circulation=  Vds =  Vds +  Vds +  Vds +  Vds
c
Each of these curves must be evaluated individually.
Along the curve AB where r = R ( l1 < R < 100) and θ ranges from 0 to 2π ,
B
2π
2π
2π
Γ
Γ
 Γ 
 Vds =  Vθ Rdθ =   2π R  Rdθ = 2π  dθ = Γ , where Vθ r=R = 2π R was used in evaluat0
0
0
A
ing the integral.
Along the curve BC where θ = 2π and r ranges from R to ε (ε << 1),
C
ε
ε
B
R
R
 Vds =  Vr dr =  (0)dr = 0 where Vr = 0 was used.
Along the curve CD, r = ε and θ ranges from 2π to θ ,
0
0
0
Γ
Γ
Γ
Γ
was used to
V
d
s
V
ε
d
θ
ε
d
θ
dθ =
(0 − 2π ) = −Γ, where Vθ r =ε =
=
=
=

 θ
 2πε

2
πε
2
π
2
π
2π
2π
2π
C
evaluate the integral.
D
Along the curve DA, θ = 0 and r ranges from ε to R.
A
R
R
D
ε
ε
 Vds =  Vr dr =  (0)dr = 0
where Vr = 0 was used to evaluate the integral. Substituting the evaluation of the four integrals into the circulation expression gives:
Circulation = Γ + 0 − Γ + 0 = 0
Problem 6.50
Show that the circulation of a free vortex for any closed path that encloses the origin is Γ.
Solution 6.50
Γ
(θ )
2π
The velocity components, in polar coordinates, for a free vortex
are
∂φ ∂  Γ 
θ=0
Vr =
= 
∂r ∂r  2π 
And
1 ∂φ 1 ∂ Γ
Γ
=
( θ) =
Vθ =
r ∂θ r ∂θ 2π
2π r
The velocity potential for a free vortex is given by φ =
iy
x
A closed curve C with radius R(0 < R < ∞ ) is defined as shown above, a circle with its center
located at the origin. This curve was chosen in order to simplify the integral in the circulation expression, which is
2π
Circulation =  Vds =  Vθ Rdθ
0
c
and
Circulation = 
Γ 
Γ 2π
Γ
dθ =
[θ ]20π

 Rdθ =

0
2π
2π
 2π R 
2π 
0
Γ
[2π − 0] = Γ,
where
2π
Γ
 Γ 
. then
Vθ r =R = 
=

 2π r r =R 2π R
=
Circulation = Γ
Problem 6.51
Potential flow against a flat plate (the figure below (a)) can be described with the stream
function
ψ = Axy
where A is a constant. This type of flow is commonly called a stagnation point flow since it
can be used to describe the flow in the vicinity of the stagnation point at O . By adding a
source of strength m at O , stagnation point flow against a flat plate with a“bump” is obtained as illustrated in the figure below(b). Determine the relationship between the bump
height, h, the constant, A, and the source strength, m.
y
O
x
(a)
y
h
x
Source
(b)
Solution 6.51
ψ = Axy +
m
A
m
θ = r 2 sin 2θ + θ
2π
2
2π
π


For the bump, the stagnation point will occur at x = 0, y = h  0 = , r = h  .
2


For the given stream function,
vr =
1 ∂ψ
m
= Ar cos 2θ +
r ∂θ
2π r
(1)
And
vθ = −
∂ψ
= Ar sin 2θ
∂r
π
The point, 0 = , r = h, will be a stagnation point if vr = 0 since vθ = 0 at this point.
2
Thus, from Eq. (1)
0 = Ah cos π +
m
2π h
or
Ah =
m
2π h
And therefore
h2 =
m
2π A
Problem 6.52
The combination of a uniform flow and a source can be used to describe flow around a
streamlined body called a half-body. Assume that a certain body has the shape of a halfm
body with a thickness of 0.5 m. If this body is placed in an airstream moving at 15 , what
s
source strength is required to simulate flow around the body?
Solution 6.52
y
U
ψ = π bU
Stagnation point
Stagnation
point
r
πb
θ
x
πb
Source
b
b
(a)
(b)
The width of half-body = 2π b so that
b=
(0.5 m)
2π
From the equation
b=
m
2πU
where m is the source strength, and therefore
m
 m  0.5 m 
m = 2πUb = 2π 15 
 = 7.50
s
 s  2π 
2
Problem 6.53
Show that if Φ1 and Φ 2 are both solutions of Laplace’s equation, the sum ( Φ1 + Φ 2 ) is also a
solution.
Solution 6.53
Laplace’s equation in Φ is
∂2Φ
∂2Φ
∂2Φ
∂x
∂y
∂z 2
+
2
+
2
=0
If Φ1 is a solution, then
∂ 2Φ1
∂ 2 Φ1
∂ 2 Φ1
∂x
∂y
∂z 2
+
2
+
2
= 0 (1)
If Φ 2 is a solution, then
∂ 2Φ 2
∂2Φ2
∂x
∂y2
+
2
+
∂ 2Φ2
∂z 2
= 0 (2)
Adding Eqs. (1) and (2)
∂ 2 Φ1
∂x2
+
∂ 2 Φ1
∂y2
+
∂ 2 Φ1
∂z 2
+
∂2Φ2
∂x2
+
∂2Φ2
∂y2
+
∂2Φ2
∂z 2
=0
And
∂2
∂2
∂2
∂x
∂y
∂z 2
( Φ1 + Φ 2 ) +
2
( Φ1 + Φ2 ) +
2
( Φ1 + Φ2 ) = 0
Therefore the sum ( Φ1 + Φ 2 ) is also a solution of Laplace’s equation.
Problem 6.54
The flow in the impeller of a centrifugal pump is modeled by the superposition of a source
and a free vortex. The impeller has an outer diameter of 0.5 m and an inner diameter of
0.3 m. At the outlet from the impeller, the flowing water has the following velocity compom
m
and tangential component 7 .
nents, relative to the impeller: radial component 2
s
s
(a) Find the strength of the source and the vortex required to model this flow.
(b) Assume that the impeller blades are shaped like the streamlines and plot an impeller
blade shape.
(c) Find the radial and tangential components of velocity at the inlet to the impeller.
Solution 6.54
y
Vθ
Vr
Outlet “O”
θ
x
(a) From source flow
λ
2π r
Vr =
m2
 m
=
π

s
 s 
λ = 2π rVr = 2π rθVr 0 = 2π (0.25 m)  2
From vortex flow
Vθ =
Γ
2π r
m
 m
Γ = 2π rVθ = 2π rθVθ 0 = 2π (0.25 m)  7  = 3.5π
s
 s 
(b) From the equation
ψ=
λ
Γ
θ−
ln r
2π
2π
The equation of any streamline is
 2πψ 
 λθ 
r = exp  −
 exp 

Γ 

 Γ 
2
Consider the streamline with r = ri = 0.1 m at θ = 0
 2πψ 
0.15 = exp  −
 exp(0)
 3.5π 
ψ =−
3.5
ln(0.15) = 3.32
2
The equation of this streamline is
2π 

 π

θ
r = exp  −3.32
 exp 
3.5π 

 3.5π 
 θ 
r = 0.15 exp 

 3.5 
theta
(deg)
0.0
20.0
40.0
60.0
80.0
100.0
102.4
r
(meters)
0.1500
0.1657
0.1831
0.2023
0.2235
0.2470
0.2500
Vane (Streamline)
r = 0.15 m
r = 0.15 m
(c) At inlet ( i ) ri = 0.15 m
λ
π
m
=
= 3.33
2π r 2π (0.15)
s
Γ
3.5π
m
=
= 11.67
Vθ =
2π r 2π (0.75)
s
Vr =
Problem 6.55
One end of a pond has a shoreline that resembles a half-body as shown in the figure below.
A vertical porous pipe is located
Pipe
5m
A
15 m
near the end of the pond so that water can be pumped out. When water is pumped at the
m3
rate of 0.008
through a 3-m-long pipe, what will be the velocity at point A?
s
Hint: Consider the flow inside a half-body.
Solution 6.55
For a half-body,
ψ = Ur sin θ +
m
θ
2π
So that
vθ = −
∂ψ
= U sin θ
∂r
And
vr =
1 ∂ψ
m
= U cos θ +
2π r
r ∂θ
Thus, at point A, θ = 0, r = 15 m and
vθ = 0
vr = VA = U +
m
2π (15)
For a flowrate of 0.06
b=
(1)
m3
m2
in a 3-m long pipe, the source strength is 0.06/3
. Since
sin
s
m
2πU
then with b = 5 m
 0.06 m 2 


3 s 
m
m

=
= 6.37 × 10 −4
U=
2π b
2π (5 m )
s
From Eq. (1)
 0.06 m 2 


3 s 
−4 m 
VA = 6.37 × 10
+
s
2π (15m )
= 8.49×10−4
m
s
Problem 6.56
ft 2
A Rankine oval is formed by combining a source-sink pair, each having a strength of 36
s
ft
and separated by a distance of 12 ft along the x axis, with a uniform velocity of 10 (in the
s
positive x direction). Determine the length and thickness of the oval.
Solution 6.56
1
l  m
2
=
+ 1
a  πUa 
(1)
2

  πUa  h 
h 1  h 
=   − 1 tan  2 
 
a 2  a 
  m a

For m = 36
πUa
m
=


(2)
ft 2
ft
, a = 6 ft, and U = 10 ,
s
s
ft 
π 10  ( 6 ft)
s

ft 2
36
s
= 5.24
Thus, length = 2l and from Eq. (1)
1
 1
2
length = 2( 6 ft) 
+ 1 = 13.1ft
 5.24 
The thickness, 2h , can be determined from Eq. (2) by trial and error. Assume value for
and compare with right-hand side of Eq. (2) (see the table below).




0.250
0.251
0.252
0.253
h
≈ 0.253
a
And thickness = 2h = 2(6ft)(0.253)=3.04ft
Thus,



−




0.269
0.262
0.256
0.250






use
h
a
Problem 6.57
A 15-mph wind flows over a Quonset hut having a radius of 12 ft and a length of 60 ft, as
shown in the figure below. The upstream pressure and temperature are equal to those inside
the Quonset hut: 14.696 psia and 70 °F. Estimate the upward force on the Quonset hut.
Find the location θ on the roof of the Quonset hut where the pressure is p∞.
V∞
p∞, T∞
p∞, T∞
θ
12 ft
Solution 6.57
The air density is found by the ideal gas law.
ρ=
lb  
in2 

14.696
144




in2  
ft 2 

P
lbm
=
= 0.075 3 , and
RT 
ft
ft ⋅ 1b 
 53.36
 (530 R)
lbm R 

ft 



ft
 mi 
V∞ =  15  1.47 s  = 22.05
mi 
s
 hr  
hr 

The velocity components for flow about a circular cylinder are given by equations:
 R2 
 R2 
Vr = V∞ cos θ 1 − 2  and Vθ = V∞ sinθ 1 + 2 


r 
r 


On the cylinder surface, r = R and Vr = 0 , V∞ = −2V∞ sin θ
Bernoulli’s equation gives the pressure distribution over the exterior of the hut,
1
1
p∞ + ρV∞ 2 = p + ρV 2 where V 2 = Vr 2 +Vθ 2 . Then
2
2
2
1
1
1
p = p∞ + ρV∞ 2 − ρ ( −2V∞ sin θ )2 = p∞ + ρV∞2 (1 − 4 sin θ )
2
2
2
The location θ on the roof of the Quonset hut where the pressure in p∞ is where 4 sin2 θ = 1
or θ = 30 or 120 .
The upward force on the Quonset hut is found by integrating the difference in pressure
about the surface.
π

1


F =   p∞ −  p∞ + ρV∞ 2 (1 − 4 sin2 θ ) LRdθ
2


0
π
π
1
1

 1 − cos 2θ  
F = −  ρV∞ 2 LR(1 − 4 sin2 )dθ = − ρV∞2 LR  1 − 4 
  dθ
2
2
2


0
0
π
1
1
1
 sin 2θ  π
F = V∞2 LR  (1 + 2 cos 2θ )dθ = ρV 2 LR − ρV∞2 LR 
 |0
4
4
2
2

0
F=
1
ρV 2 LRπ
4
Inserting the numerical values gives
2
 sec 2 ⋅ lb 
1
lbm 
ft 
F =  0.075 3   22.05  (60 ft)(12 ft)π 
 32.2 lbm ⋅ ft 
4
s
ft  


F = 640 lb
Upward Force = 640lb
Problem 6.58
An ideal fluid flows past an infinitely long, semicircular “hump” located along a plane
boundary, as shown in the figure below. Far from the hump, the velocity field is uniform,
and the pressure is p0. (a) Determine expressions for the maximum and minimum values of
the pressure along the hump, and indicate where these points are located. Express your
answer in terms of ρ, U , and p0. (b) If the solid surface is the ψ = 0 streamline, determine the
equation of the streamline passing through the point θ =
π
2
, r = 2a.
U, p0
r
a
θ
Solution 6.58
(a) On the surface of the hump,
1
(1)
ps = p0 + ρU 2 (1 − 4 sin2 θ )
2
The maximum pressure occurs where sinθ = 0, or at θ = 0, π , and at these points
1
(at θ = 0 and π )
ps (max) = p0 + ρU 2
2
The minimum pressure occurs where sinθ = 1, or at θ =
π
2
, and at this point

π
 at θ = 
2

(b) for uniform flow in the negative x-direction,
ps (min) = p0 −

3
ρU 2
2
a2 
ψ = −Ur 1 − 2  sin θ
 r 
(refer to discussion associated with the derivation of Eq. (1)).
π
At θ = , r = 2a
2

a2  π
3
ψ = −2aU 1 −
sin = − aU
2

2
2
 (2a ) 
And thus the equation of the streamline passing through this point is
 a2 
3
− aU = −Ur 1 − 2  sin θ
 r 
2


or
3 r  a2 
1 −  sin θ = 1
2 a  r 2 
Problem 6.59
Assume that the flow around the long circular cylinder of the figure below is nonviscous
and incompressible. Two pressures, p1 and p2, are measured on the surface of the cylinder,
as illustrated. It is proposed that the free-stream velocity, U , can be related to the pressure
difference Δp = p1 − p2 by the equation
Δp
p
where p is the fluid density. Determine the value of the constant C. Neglect body forces.
U =C
U
y
p2
p1
a
60°
x
Solution 6.59
Since p1 is the stagnation pressure,
p1
γ
=
p2 V22
+
γ 2g
(1)
where
V2 = vθ s = − 2U sin θ
For θ = 60 ,
V22 = ( −2U sin60 )2 = 3U 2
Thus, from Eq. (1)
p1 − p2 =
1γ 2 1
3
V2 = ρ (3U 2 ) = ρU 2
2g
2
2
So that U =
2
3
And therefore
C=
2
3
p1 − p2
ρ
=
2 Δp
3 ρ
Problem 6.60
Consider the steady potential flow around the circular cylinder as shown in the figure below. On a plot show the variation of the magnitude of the dimensionless fluid velocity,
V / U , along the positive y axis. At what distance, y / a (along the y axis), is the velocity within 1% of the free-stream velocity?
U
2U
Ψ=0
r
θ
a
Solution 6.60
The velocity components are given by Eqs. (6.114) and (6.115)
 a2 
vr = U  1 − 2  cos θ
 r 


 a2 
vθ = −U 1 + 2  sin θ
 r 


With U as the free-stream velocity and a as the cylinder radius. Along the positive y axis,
θ=
π
2
;
cos θ = 0
sin θ = 1
therefore
2
V −vθ   a  
=
= 1 +   
  y 
U U


y
≥1
a
r=y
The following plot was made using Microsoft® Excel®
Dimensionless Velocity Above
the Crest of a Circular Cylinder
y/a
V/U
The “1%” point is found by
2
a
V
y
= 1.01 = 1 +    = 10
U
a
 y
Problem 6.61
A highway sign has the shape of a half-cylinder with radius 1.0 m and length 25.0 m. A
rearward 10 km/hr wind is blowing over the sign as shown in the figure below.
y
Wind
p
θ
R
2m
x
V∞
Message here
Length Perpendicular
to paper = 25 m
The pressure on the flat face is constant and equal to the pressure at the corner. Find the
force on the sign. Neglect end effects.
Solution 6.61
Sign having shape of half-cylinder, as shown in the figure below.
y
π –θ
Wind
p
π –θ
τw
R
V∞
θ
2m
x
Length perpendicular
to paper = 25 m
The directions of positive lift and drag are the positive y and x direction, respectively.
dA = bR dθ ,
where b is the sign length and R is the radius. The drag is
D=
2π
0
( − p cos θ + τ w sinθ ) bR dθ .
Recognizing that the drag on the top half equals the drag on the bottom half gives
π
D = 2 ( − p cosθ + τ w sinθ ) bRdθ = Dp + Dτ .
0
For this problem, we assume that the viscous shear force is negligible. We evaluate the pressure integral in two “pieces”, one for the flat face and one for the cylindrical portion. On
the flat face, the pressure is given by

 π 
p − p∞ = 4.63 1 − 4 sin2    = 4.63 ( −3 ) .
 2 

Thus,
D = 2
π 2
0
+2 
−  p∞ + 4.63 ( −3)  cos θ bR dθ
(
π
)
−  p∞ + 4.63 1 − 4 sin2 θ  cos θ bR dθ .


π 2
Integrating, we obtain
π 2
D = −2bR ( p∞ − 13.9 ) [sin θ ]0
π

 sin3 θ  
−2bR  p∞ sin θ + 4.63sin θ − 18.5 
.
 3  



 π 2
Substituting limits gives
(
)
(
D = 2bR 12.36 N m 2 = 2 ( 25 m )(1m ) 12.36 N m 2
)
D = 618 N.
Next find the lift. Recognizing that dA = bR dθ and that there is no lift component on the
flat face, we have
L=
3π 2
π 2
( − p sin θ − τ w cos θ )bR dθ ,
Substituting for p and assuming τ w = 0 give
L=
3π 2
π 2
(
)
−  p∞ + 4.63 1 − 4 sin 2 θ  sin θ bR dθ


We now use the trigonometric identity
sin 2 θ = 1 − cos 2 θ
in the integral. The lift equation then is
L = bR 
3π 2
π 2
(
)
−  p∞ + 4.63 −3 + 4 cos 2 θ  sin θ dθ


Integrating yields
3π 2

18.52 cos3 θ 
L = bR ( p∞ − 13.9 ) cos θ +

3

π 2
which reduces to
L = ( 0 ) bR ,
so
L = 0.
Comments The zero lift should not be surprising, because the vertical component of the
pressure (or shear stress) at any point on the top half of the sign is exactly balanced by an
equal but oppositely directed vertical component of the pressure (or shear stress on the bottom half). Unlike that of an airfoil, this object’s drag is dominated by pressure force rather
than by shear stress force, as we will see in later chapters.
Problem 6.62
Air at 25 °C flows normal to the axis of an infinitely long cylinder of 1.0 m radius. The cylrad
km
, and the approach velocity is 100
. Find the maximum and
inder is rotating at 10
s
hr
minimum pressures on the cylinder surface. Assume potential flow.
Solution 6.62
For 25 C , ρ = 1.19
kg
m3
. Applying Bernoulli’s equation to the air gives
p0 = stagnation pressure = p∞ +
Vθ = −
∂Ψ
,
∂r
Vr =
ρV∞2
2
= p+
ρV 2
2
, where V 2 = Vθ2 +Vr2 and
1 ∂ψ
,
r ∂θ
And
ψ = V∞ r sin θ −V∞ R 2
sin θ Γ
+
ln r,
r
2π
Then
Vθ r =R = −

R2
∂ψ
Γ 
= − V∞ sin θ + V∞ sin θ 2 +

2π r 
∂r r = R
r

= −2V∞ sin θ −
r =R
Γ
,
2π R
and
Vr =
1 ∂ψ
cos θ 

= V∞r cos θ −V∞R2
= V∞R cos θ −V∞R cos θ = 0
r ∂θ r =R 
r  r =R
On the cylinder surface,
2
p = p∞ + ρ
V∞ 2
V 2 ρ
V2
Γ 
.
−ρ
= p∞ + ρ ∞ −  −2V∞ sin θ −
2
2
2
2
2π R 
The location of the minimum and maximum pressures is found by setting
ing for θ . This gives
∂p
Γ 

= − ρ  2V∞ sin θ +
 (2V∞ cos θ ) = 0 ,
∂θ
2π R 

cos θ = 0 → θ =
π
2
,
3π
,
2
and
2V∞ sin θ +


Γ
Γ
= 0 → θ = sin −1  −

2π R
 4π RV∞ 
∂p
= 0 and solv∂θ
Since Γ = 2π RU = 2π R2ω ,

 rad  
(1.0 m) 10




Rω
s  

−1 
=
sin
−
= sin−1( −0.13) = −10.37 , 190.37
θ = sin−1  −

m 


 2V∞ 
2  27.78


s  


Then
p∞ +
ρV∞2
2
(1.19
= 101300 Pa+
kg
m
)(27.78 )2
3
s = 101760 Pa ,
m
2
ρ

 ρ
pR =  p∞ + V∞2  − (2V sin θ + Rω )2 ,
2

 2
kg 

 1.19 3 
2
m  [(2)(27.78)(1) + (1.0)(10)]2 m = 99200 Pa
p π = 101760 Pa − 
R1
2
s2
2
kg 

 1.19 3 
2
m  [(2)(27.78)( −1) + (1.0)(10)]2 m = 100500 Pa
p 3π = 101760 Pa − 
R1
2
s2
2
kg 

1.19 3 
2
m  [(2)(27.78)( −0.18) + (1.0)(10)]2 m = 101760 Pa
pR1−10.37 = 101760 Pa − 
2
s2
kg 

1.19 3 
2
m  [(2)(27.78)( −0.18) + (1.0)(10)]2 m = 101760 Pa .
pR,190.37 = 101760 Pa − 
2
s2
Then
pmax = 101760 Pa at θ = −10.37 ,
θ = 190.37
pmin = 99.202 kPa at θ =
π
2
Problem 6.63
The velocity potential for a cylinder (the figure below) rotating in a uniform stream of
fluid is

a2 
φ = Ur 1 + 2  cos θ +
Γ
θ
2π
 r 
where Γ is the circulation. For what value of the circulation will the stagnation point be located at: (a) point A; (b) point B?
y
U
r
θ
A
a
x
B
Solution 6.63
Γ
(1)
4πUa
At point A, θstag = 0 and it follows that Γ = 0
(a) sin θ stag =
(b) At point B, θ stag =
Γ = 4πUa sin
3π
= − 4πUa
2
3π
, and from Eq. (1)
2
Problem 6.64
Determine the magnitude of the total force developed by the two rotating cylinders on the
Flettner “rotor-ship” due to the Magnus effect. Assume a wind-speed relative to the ship of
(a) 10 mph and (b) 30 mph. Each cylinder has a diameter of 9 ft, a length of 50 ft , and rotates at 750 rev/min. Use the equation Fy = − ρU Γ and calculate the circulation by assuming
the air sticks to the rotating cylinders. Note: This calculated force is at right angles to the
direction of the wind and it is the component of this force in the direction of motion of the
ship that gives the propulsive thrust. Also, due to viscous effects, the actual propulsive
thrust will be smaller than that calculated from the equation Fy = − ρU Γ , which is based on
inviscid flow theory.
Solution 6.64
Fy = − ρU Γ (force per unit length)
Γ=
 Vds
On the cylinder surface
V = rω eˆ θ and ds = rdθ eˆ θ
So that
Γ
2π
0
rev   rad   1min 
ft 2

=
π
( rω )( rdθ )eˆ θ ⋅ eˆ θ = 2π r 2ω = (2π )(4.5 ft)2  750
2
9990



min  
rev   60 s 
s

And
slugs  
ft 2 

Fy = −  0.00238
9990

U = −23.8U

s 
ft3  

(a) For a cylinder with length = 50 ft and number of cylinders = 2
and wind speed = 10 mp 4,
lb ⋅ s   mi  
ft   l hr 

Fy =  23.8 2  10
 (50 ft)(2) = 34900 lb
  5280

mi   3600 s 
ft   hr  

(b) At 30 mph
Fy = 3 × ( Fy @10 mph) = 105000 lb
Problem 6.65
Consider the possibility of using two rotating cylinders to replace the conventional wings
km
through the Standard Atmoson an airplane for lift. Consider an airplane flying at 700
hr
phere at 10,000 m. Each “wing cylinder” has a 3.0-m radius. The surface velocity of each
km
cylinder is 28
. Find the length of each wing to develop a lift of 1550 kN. Assume pohr
tential flow and neglect end effects.
Solution 6.65
At 10,000 m , the density is 0.4135
kg
m3
. The equation for lift and circulation about a rotat-
ing cylinder are
Γ = 2π RU , where U =
28000 m
m
= 7.78
3600 s
s
The lift L per unit length is
L
kg   700000 m 
m
kg


= ρV∞ Γ = ρV∞ 2π RU =  0.4135 3  
 2π (3 m)  7.78  = 11788 2
s 
m   3600 s 
s


The length of each wing is found from
L = 1550 kN=1550000 N or
kg ⋅ m 

 
 1550000 N   1550000

s2 
=
=
 11787 kg   11787 kg 
s2  
s2


= 131.5 m
Problem 6.66
Air at 25 °C flows normal to the axis of an infinitely long cylinder of 1.0 m radius. The apkm
proach velocity is 100
. Find the rotational velocity of the cylinder so that a single staghr
nation point occurs on the lower-most part of the cylinder. Assume potential flow.
Solution 6.66
Equation Vθ )r =R = −2V∞ sin θ −
Γ
gives the velocity at the surface of the cylinder.
2π R
The stagnation point θ = −90 is found by setting Vθ = 0
so
Γ
Γ
and 2V∞ =
2π R
2π R
2π RU
where Γ = 2π RU . Then 2V∞ =
, so
2π R
0 = −2V∞ sin( −90 ) −
m
 100000 m 
U = 2V∞ = 2 
 = 55.56
s
 3600 sec 
m
U = rω,
55.56
= (1.0 m)ω
sec
rad
And ω = 55.56
.
sec
Rotational speed is 55.56
rad
in clockwise direction.
sec
Problem 6.67
Determine the shearing stress for an incompressible Newtonian fluid with a velocity distri.
bution of =
−
+
−
Solution 6.67
The shearing stress for an incompressible* Newtonian fluid is
τ
=τ
∂
∂ 
= μ + 
∂ 
∂
τ
=τ
∂ ∂ 
= μ +

∂ 
∂
τ
=τ
∂ 
∂
= μ
+ 
∂ 
∂
For the given velocity vector,
=
−
=
−
=
Thus,
τ =μ
τ
=μ
+
=
+
=
+
=
μ
and
τ
=μ
*Note: for the given velocity
∂
∂ ∂
∇⋅ =
+ +
=
−
∂ ∂
∂
+
Hence,
∇i = , the flow is incompressible.
−
+
=
Problem 6.68
The two-dimensional velocity field for an incompressible Newtonian fluid is described by
the relationship
V = (12 xy 2 − 6 x 3 )ˆ + (18x 2 y − 4 y3 )ˆ
where the velocity has units of m/s when x and y are in meters. Determine the stresses σ xx ,
σ yy , and τ xy at the point x = 0.5 m,y = 1.0 m if pressure at this point is 6 kPa and the fluid is
glycerin at 20 °C. Show these stresses on a sketch.
Solution 6.68
∂u
∂x
∂v
τ yy = − p + 2 μ
∂y
τ xx = − p + 2 μ
 ∂u ∂v 
+ 
 ∂y ∂x 
τ xy = μ 
For the given velocity distribution, with x = 0.5 m and y = 1.0 m :
∂u
1
= 12 y2 − 18x 2 = 12(1.0)2 − 18(0.5)2 = 7.50 ,
∂x
s
∂u
1
= 24xy = 24(0.5)(1.0) = 12.0 ,
∂y
s
∂v
1
= 36xy = 36(0.5)(1.0) = 18.0
∂x
s
∂v
1
= 18x2 − 12 y2 = 18(0.5)2 − 12(1.0)2 = −7.50
∂y
s
Thus, for p = 6 × 103
N
m
2
and μ = 1.50
N ⋅s
m2
,
σ xx = −6 ×103
N ⋅ s 
1

+ 2 1.50 2  7.50  = −5.98kPa
s
m
m 

σ yy = −6 ×103
N ⋅ s 
1

+ 2 1.50
−7.50  = −6.02 kPa
2 
s
m
m 



τ xy = 1.50
N
2
N
2
N ⋅s 
1
1
12.0 + 18.0  = 45.0 Pa
2 
s
s
m 
6.02 kPa
y
45.0 Pa
5.98 kPa
x
Problem 6.69
The velocity of a fluid particle moving along a horizontal streamline that coincides with the
x axis in a plane, two-dimensional, incompressible flow field was experimentally found to
be described by the equation u = x 2 . Along this streamline, determine an expression for
(a) the rate of change of the v component of velocity with respect to y , (b) the acceleration
of the particle, and (c) the pressure gradient in the x direction. The fluid is Newtonian.
Solution 6.69
(a) From the continuity equation,
∂u ∂v
+
=0
∂x ∂y
So that with u = x 2
∂v
∂u
=−
= −2x
∂y
∂x
(1)
Also, Eq. (1) can be integrated with respect to y obtain
 dv =  −2xdy
or
v = −2xy + f ( x )
Since the x-axis is a streamline, v = 0 along this axis and therefore f ( x ) = 0 so that
v = −2xy
(b) ax = u
ay = u
∂u
∂u
+v
= ( x2 )(2x ) + ( −2xy )(0) = 2x3
∂x
∂y
∂v
∂v
+ v = ( x2 )( −2 y ) + ( −2xy )( −2x ) = 2x2 y
∂x
∂y
Along x-axis , y = 0, and therefore a y = 0 . Thus,
a = 2x3ˆ
(c) With g x = 0 ,
ax = −
1 ∂p μ  ∂ 2 u ∂ 2 u 
+ 
+

ρ ∂x ρ  ∂x 2 ∂y2 
So that
2 x3 = −
1 ∂p μ
+ (2 + 0)
ρ ∂x ρ
and
∂p
= 2 μ − 2 ρ x3
∂x
Problem 6.70
“Stokes’s first problem” involves the instantaneous acceleration at time t = 0 of a flat plate
to a constant velocity U0 while in contact with a “semi-infinite,” static fluid as shown in the
figure below. For a constant fluid density and viscosity, the simplified Navier–Stokes equation is
∂u
∂2u
=v 2
∂t
∂y
where u is the fluid velocity in the x or velocity U0 direction and y is a coordinate normal to
the plate. Find the appropriate boundary conditions and initial conditions for this problem
u
= f ( y, t ).
and then solve the differential equation to determine the velocity distribution
U0
y
Hint: Assume that f is a function of a single variable η where η =
vt .
2
“Semi-infinite” fluid
t ≤ 0, u = 0
u ( y, t )
y
Plate
x
U0
Solution 6.70
There are two boundary conditions for t > 0
y = 0, u = U 0
y =∞, u = 0
There is one initial condition for y > 0
t = 0, u = 0
Following the hint,
u
y
= f ( y, t ) = f (η ) where η =
U0
2 νt
Evaluating the derivatives
U
U η
U η
∂u
y df
=− 0
= − 0 f ′(η ) = − 0 f ′
2 2 ν t dη
2 t
2 t
∂t
∂u U 0
f′
=
∂y 2 ν t
∂ 2u
∂y
2
=
U0
f ′′
4ν t
Substituting into the differential equation, collecting terms and assuming t ≠ 0 so that t may
be cancelled gives
f ′′ + 2η f ′ = 0
Dividing by f ′ gives
f ′′ d (ln f ′)
=
= −2η
f′
dη
Integrating gives
ln f ′ = −η 2 + C = −η 2 + ln A
Taking the antilogarithm
f ′ = Ae−η
2
and
2
η
u
= f (η ) = A e −ξ dξ + B
0
U0
The boundary conditions give
u(0, t ) = U 0  f (0) = 1
u( ∞, t ) = u( y,0) = 0  f ( ∞ ) = 0
so
2

∞
π 
2
+ 1  A = −
0 = U 0   e−ξ dξ + 1 = U 0  A
0


π
 2

The solution is
u
2 y /2 ν t −ξ 2
 y 
e d ξ = 1 − erf 
= 1−


0
U0
π
 2 νt 
Problem 6.71
Oil (SAE 30) at 15.6 °C flows steadily between fixed, horizontal, parallel plates. The preskPa
sure drop per unit length along the channel is 30
, and the distance between the plates is
m
4mm. The flow is laminar. Determine: (a) the volume rate of flow (per meter of width),
(b) the magnitude and direction of the shearing stress acting on the bottom plate, and
(c) the velocity along the centerline of the channel.
Solution 6.71
2h3 Δ p
(a) q =
3μ
For h =
4 mm
N ⋅s
Δp
N
= 2 × 10 −3 m , μ = 0.38 2 , and
= 30 × 103 3 ,
2
m
m
N 

2(2 ×10−3 m)3  30 ×103 3 
2
m  = 4.21×10−4 m

q=
N⋅s 
s

3  0.38
2 

m 
 ∂u ∂v 
(b) τ yx = μ  + 
 ∂y ∂x 
Since
u=
1 ∂p 2 2
(y − h )
2 μ ∂x
and v = 0
If follows that
∂u
1
∂p
=
(2 y )
∂y 2 μ
∂x
∂v
=0
∂x
and therefore
τ yx = ( y )
∂p
∂x
At the bottom plate, y = − h , and since
τ yx =
∂p
Δp
=− ,
∂x
Δp
N 

( h ) =  30 × 103 3  (2 × 10 −3 m)
l

m 
=60
N
m2
acting in the direction of flow
2

−4 m 
 4.21× 10

s 
3
3  q  3 
m
(c) umax = V =   =
=
0.158
2
2  2 h  2 (2)(2 × 10 −3 m)
s
Problem 6.72
Two fixed, horizontal, parallel plates are spaced 0.4 in. apart. A viscous liquid
ft
lb ⋅ s
( μ = 8 × 10 −3 2 , SG = 0.9) flows between the plates with a mean velocity of 0.5 . The
s
ft
flow is laminar. Determine the pressure drop per unit length in the direction of flow. What
is the maximum velocity in the channel?
Solution 6.72
h2 Δp
V=
3μ
lb ⋅ s  
ft 

3  8 × 10 −3 2   0.5 
lb ⋅ s
Δp 3μV
s
ft  
Thus,
= 2 = 
= 43.2 2 per ft


h
ft
 0.2 in. 
 in. 
 12

ft 

3
3
ft 
ft
U max = V =  0.5  = 0.75
2
2
s
s
Problem 6.73
A liquid of constant density ρ and constant viscosity μ flows down a wide, long inclined flat
plate. The plate makes an angle θ with the horizontal. The velocity components do not
change in the direction of the plate, and the fluid depth, h, normal to the plate is constant.
There is negligible shear stress by the air on the fluid. Find the velocity profile u( y ), where u
is the velocity parallel to the plate and y is measured perpendicular to the plate. Write an
expression for the volume flowrate per unit width of the plate.
Solution 6.73
h
y
θ
x
Consider a force balance on an element of fluid as shown below.
SW = ρgδ xδ yδ z
Δy
τ+
Δx
∂x
δy
∂y
element thickness
Δz ⊥ to paper
τ
Since the velocities do not change in the plate direction, there is no acceleration in this direction (or in any other direction). The motion is determined by the balance between the
gravity force and shear stresses in the x-direction,
δ Fstress + δ Fgravity = 0
Substituting expressions for each gives


∂τ
τ + δ y  δ xδ z − τ δ xδ z + ρ δ xδ yδ z g sin θ = 0
∂y 

Simplifying gives
∂τ
= − ρ g sin θ
∂y
Since τ is a function of velocity which does not change with x , the stress is given by
du
τ =μ
dy
so μ
d 2u
dy2
= − ρ g sin θ
Integrating twice gives
u=−
ρ g sinθ 2
y + C1y + C2.
2μ
The boundary conditions are
u = 0 at y = 0 and
du τ
= = 0 at y = h .
dy μ
These conditions give
0 = C 2 and 0 = −
Solving gives C 2 = 0 and C1 =
ρ g sin θ
h + C1.
μ
ρ g sin θ h
μ
ρ g sin θ h2  y 1  y  
 −    .
μ
 h 2  h  
2
The velocity profile is then u =
The flowrate per unit width of the plate is calculated by
h
1
 y
Q =  udy = h  ud   .
0
0
h
Substituting for u gives
ρ g sin θ h2 ( h ) 1  y 1  y    y 
0  h − 2  h   d  h  ,
μ


2
Q=
3 1
ρ g sin θ h3  1  y  1  y  
Q=
   −   
μ
 2  h  6  h  
2
or
Q=
ρ g sin θ h3
3μ
0
Problem 6.74
We will see in Chapter 8 that the pressure drop in fully developed pipe flow is sometimes
computed with the aid of a friction factor, defined by
f =
Δp
D
1
ρV 2
2
where V is the average velocity and is the length of pipe over which Δp occurs. For laminar
fully developed flow, f can be evaluated from
f =
C
Re
where C is a constant and Re is the Reynolds number, given by Re = ρVD / μ . Find the
constant C for a parallel plate channel. Take D as 2h.
Solution 6.74
From Eq. (6.137), for flow between parallel plates
V=
h2 Δp
3μ
Then
f =
=
Δp
1 ρV
2
D
2
=
Δp
 2 
1 ρ  h Δp 
2  3μ 


(4hΔp ) × (3μ )
(
( ρV ) h2 Δp
So C = 24
2h
)
=
2
=
12 hΔp μ
(
( ρV ) h h Δp
2 × 2 × h × Δp
=
ρ ×V ×V ×
)
=
4 × h × Δp
 h2 Δp 
ρ ×V × 
 ×
 3μ 
24 μ
24
=
ρV (2h ) Re
Problem 6.75
A massive, precisely machined, 6-ft-diameter granite sphere rests on a 4-ft-diameter cylindrical pedestal as shown in the figure below.
6 ft
0.005 in.
4 in.
4 ft
p = 8 psi
Pump
When the pump is turned on and the water pressure within the pedestal reaches 8 psi, the
sphere rises off the pedestal, creating a 0.005-in. gap through which the water flows. The
sphere can then be rotated about any axis with minimal friction. (a) Estimate the pump
flowrate, Q0, required to accomplish this. Assume the flow in the gap between the sphere
and the pedestal is essentially viscous flow between fixed, parallel plates. (b) Describe what
would happen if the pump flowrate were increased to 2Q0 .
Solution 6.75
2h3Δp
flowrate
where q =
(a) q =
(1)
3μ l
unit width
h=
0.005 in.
= 0.0025 in.=2.08 ×10 −4 ft
2
8 psi
Q0
0 psi
h
0.005 in.
h
4 in.
width = π (4 ft)
2
 lb   144 in. 
(2)(2.08 × 10 −4 ft)  8



3
 in.2   ft 2 
−4 ft
q=
= 8.86 × 10
per unit width
s


lb × s   4 in. 

3  2.34 ×10 −5 2  

ft   12 in. 

 ft 
Thus,
2

ft3
gallons
−4 ft 
Q0 =  8.86 ×10
(4.98
)
 (4π ft) = 0.0111

s 
s
min

(b) Since 8 psi supports the sphere, it is expected that this pressure remains approximately. the same as the flowrate increases. To maintain this pressure, the distance h
would have to increase as Q0 (or q ) is increased.
Thus, from Eq. (1)
3
qnew  hnew 
=

qold  hold 
3
h 
2 =  new 
 hold 
hnew
1
= (2) 3 (0.0025 in.) = 0.00315 in.
Thus, the gap width would increase to approximately 0.00630 in.
Problem 6.76
The bearing shown in the figure below consists of two parallel discs of radius h separated
from each other by a small distance h ( h R ). Thelower disc is made of a porous material,
and an incompressible, viscous fluid is pumped through it and into the gap, filling the space
between the discs completely. The pores of the lower disc are closely spaced, so that the velocity of the fluid as it leaves the surface of the porous disc may be regarded as a uniform
value w0 , that is, small compared to the mean radial velocity. Find the radial velocity u and
the load L that the bearing can support as a function of w0 , μ, R, and h. The inertia terms
are negligible, the circumferential velocity is zero (Vθ = 0), and angular symmetry exists
(∂ ∂θ = 0 ).
L
R
h
Flow
Solution 6.76
Set up a cylindrical polar coordinate system with the z coordinate being the axis of the inflow pipe, the r coordinate radially outward, and θ the rotational coordinate.
The governing equations for the flow are the Navier–Stokes Equations in cylindrical coordinates, which can be found in Chapter 6. The following simplifications apply for the given
flow:
The flow is steady →
→ Vθ = 0
∂( )
∂( )
= 0; The flow is axisymmetric →
= 0; There is no swirl
∂t
∂θ
The equations can be further simplified by making the following assumptions:
∂p
∂z
∂p
so
∂r
p ≈ p( r ); Vx ≈ 0
Using all simplifications, the N–S equations reduce to only the r-component in the form
0=−
d 2V
dp
+ μ 2r
dr
dz
Integrating twice
Vr =
1 dp 2
z + C1z + C2
2 μ dr
Vz (0) = Vz ( h ) = 0
The boundary conditions are
so the final velocity profile is
1 dp 2
( z − zh )
2 μ dr
u = Vz =
Now apply the integral continuity equation to the control volume shown
r
w0
h
h
1 dp 2
π rh3 dp
( z − zh )dz  w0 (π r 2 ) = −
2 μ dr
6 μ dr
0
w0 (π r ) = 2π r  u dz = 2π r 
2
0
Solving for
dp
dr
6 μw
dp
=− 30 r
dr
h
Substituting into the velocity profile
u = Vr =
3w0 r
h
3
( xh − x 2 )
Now we can find the pressure
p=
6 μw
3μ w
dp
dr = −  3 0 r dr = − 3 0 r 2 + C
dr
h
h
The pressure is 0 (gage) at r = R so C =
3μ w0
h
3
R2
and
p=
3μ w0
h
3
(R2 − r2 )
The load that is supported by the bearing is equal to the total pressure force on the bearing
disc
R
 R 2r 2 r 4 
R 3μ w
6
3π μw0R 4
π
μ
w
2
2
0
0

( R − r )2π r dr =
−  =
L =  p dA = 
3
3  2
4
2 h3
h
h
0

0
Problem 6.77
A Bingham plastic is a fluid in which the stress τ is related to the rate of strain
du
by
dy
du
,
dy
where τ 0 and μ are constants. Consider the flow of a Bingham plastic between two fixed,
du
dp
horizontal, infinitely wide, flat plates. For fully developed flow with
=
= 0 and
dx
dx
constant, find u( y ).
τ = τ0 + μ
Solution 6.77
This problem is similar to the flow of a Newtonian fluid. In this case, however, we cannot
simplify the Navier–Stokes equations to obtain the equation of motion since the Navier–
Stokes equations are for a Newtonian fluid. Therefore, we must start with a force balance
on an element of fluid.
ρz + Δz Δx
τz + Δz Δx
Δx
ρx Δz
Δz
ρgΔxΔz
τz Δx
z
ρx + Δx Δz
2Z
ρz Δx
x
The element of fluid has a depth of one (1) unit. A force balance in the x-direction gives
 Fx = 0
→
+
Or px Δz − px +Δx Δz + τ z +Δz Δx − τ z Δx = 0
Dividing by ΔxΔz gives
τ z +Δz − τ z
Δz
=
pz +Δx − px
Δx
Letting Δx → 0 and Δz → 0 (noting τ ≠ f ( x ))
dτ ∂p
=
dz ∂x
(1)
A force balance in the z- direction gives
+↑
 Fz = 0
Or pz Δx − pz +Δz Δx − ρ g ΔxΔz = 0
Dividing by ΔxΔz and letting Δz → 0 gives
∂p
= −ρ g = γ .
∂z
Let the pressure at z = 0 be p0 and integrate from z = 0 to z = 2Z
With 2Z, the plate separation
P
z
P dp = − ρ g 0 dz
0
p − p0 = − ρ gz or p = p0 − ρ gz
∂p ∂p0
,
=
∂x ∂x
Then
And Eq. (1) is
dτ dp0
.
=
dz dx
Substituting τ = τ 0 + μ
μ
d 2u
2
=
du
gives
dz
dp0
dx
dz
Integrating gives
u=
1 dp0 2
z + Cz + C2 .
2 μ dx
The boundary conditions
z = 0, u = 0
give
and
z = 2Z, u = 0
0 = C2
0=
1 dp0
(4 z 2 ) + 2C1Z
2 μ dx
or
1 dp0
z(2Z − z ).
2 μ dx
(2)
Then u = −
 Z   dp 
C1 = −    0 
 μ   dx 
For τ > τ 0 . Denote the value of z at which τ = ±τ 0 by z 0
Then
τ =μ
du
1 dp0
(2Z − 2 z )
=−
dz
2 μ dx
or
±τ 0 = −
1 dp0
μτ 0
( Z − z0 ) or z0 = Z ±
μ dx
 dp0 


 dx 
So the velocity profile is
u=−
1 dp0
z(2Z − z ) for
2 μ dx
z<Z+
μτ 0
 dp0 
 dx 


and
z≥Z−
μτ 0
 dp0 
 dx 


When z = z0,
u=−
1 dp0
μτ 0
μτ 0
z0 (2Z − z0 ) for z > Z +
and z < Z −
2 μ dx
 dp0 
 dp0 
 dx 
 dx 




μτ 0
μτ 0
with z0 = Z +
or z0 = Z −
 dp0 
 dp0 
 dx 
 dx 




Problem 6.78
Two horizontal, infinite, parallel plates are spaced a distance b apart. A viscous liquid is
contained between the plates. The bottom plate is fixed, and the upper plate moves parallel
to the bottom plate with a velocity U . Because of the no-slip boundary condition, the liquid
motion is caused by the liquid being dragged along by the moving boundary. There is no
pressure gradient in the direction of flow. Note that this is a so-called simple Couette flow.
(a) Start with the Navier–Stokes equations and determine the velocity distribution between
the plates. (b) Determine an expression for the flowrate passing between the plates (for a
unit width). Express your answer in terms of b and U .
Solution 6.78
U
u
b
y
x
Fixed plate
(a) For steady flow with v = w = 0, it follows that the Navier–Stokes equations reduce to
(in direction of flow)
0=−
 ∂2u 
∂p
+μ 2 
 ∂y 
∂x


Thus, for zero pressure gradient
 ∂2u 
 2  = 0
 ∂y 
So that u = C1 y + C 2
At y = 0 u = 0 and it follows that C 2 = 0. Similarly,
At y = b u = U and C1 =
Therefore u =
U
b
U
y
b
U b
U y2
(b) q =  u(1)dy =  ydy =
0
b 0
b 2
b
b
where q is the flowrate per unit width.
=
0
Ub
2
Problem 6.79
An incompressible, viscous fluid is placed between horizontal, infinite, parallel plates as is
shown in the figure below. The two plates move in opposite directions with constant velocities, U1 and U2, as shown. The pressure gradient in the x direction is zero, and the only body
force is due to the fluid weight. Use the Navier–Stokes equations to derive an expression for
the velocity distribution between the plates. Assume laminar flow.
U1
b
y
x
U2
Solution 6.79
U1
v=0
u
g
b
w=0
y
x
U2
For the specified conditions, v = 0, w = 0,
∂P
= 0, and g x = 0 , so that the x - component of
∂x
the Navier–Stokes equations reduces to
d 2u
dy2
=0
(1)
Integration of Eq. (1) yields
u = C1 y + C 2
(2)
For y = 0, u = −U 2 and therefore from Eq. (2)
C 2 = −U 2
For y = b, u = U so that
U1 = C1b − U 2
or
C1 =
U1 +U2
b
Thus,
 U +U 2 
u= 1
 y −U 2
b


Problem 6.80
Two immiscible, incompressible, viscous fluids having the same densities but different viscosities are contained between two infinite, horizontal, parallel plates (the figure below).
The bottom plate is fixed, and the upper plate moves with a constant velocity U . Determine
the velocity at the interface. Express your answer in terms of U , μ1, and μ 2 . The motion of
the fluid is caused entirely by the movement of the upper plate; that is, there is no pressure
gradient in the x direction. The fluid velocity and shearing stress are continuous across the
interface between the two fluids. Assume laminar flow.
U
ρ , μ1
h
ρ, μ 2
y
h
x
Fixed
plate
Solution 6.80
∂p
= 0, and g x = 0 , so that the x-component of the
∂x
Navier–Stokes equations for either the upper or lower lager reduces to
For the specified conditions, v = 0, w = 0,
d 2u
dy2
=0
(1)
Integration of Eq. (1) yields
u = Ay + B
which gives the velocity distribution in either layer. In the upper layer at y = 2 h, u = U so
that
B1 = U − A1 (2 h )
where the subscript 1 refers to the upper layer . For the lower layer at y = 0, u = 0 so that
B2 = 0
where the subscript 2 refers to the lower layer. Thus,
u1 = A1 ( y − 2 h ) + U
And
u2 = A2 y
At y = h , u1 = u2 so that
A1 ( h − 2 h ) + U = A2 h
or
A2 = −A +
U
h
(2)
Since the velocity distribution is linear in each layer the shearing stress
 ∂u ∂v 
du
+ =μ
dy
 ∂y ∂x 
τ xy = μ 
is constant throughout each layer. For the upper layer
τ 1 = μ1A1
and for the lower layer
τ 2 = μ 2 A2
At the interface τ 1 = τ 2 so that
μ1A1 = μ 2 A2
or
A1 μ2
=
A2 μ1
(3)
Substitution of Eq. (3) into Eq. (2) yields
A2 = −
μ2
U
A2 +
μ1
h
or
U
A2 = h
μ
1+ 2
μ1
Thus, velocity at the interface is
u2 ( y = h ) = A2 h =
U
1+
μ2
μ1
Problem 6.81
The disc shown in the figure below has a diameter D = 1 m and a rotational speed of
1800 rpm. It is positioned 4 mm from a solid boundary. The gap is filled with 15 °C SAE
10 W oil. Find the torque required to overcome the frictional resistance of the oil on the disc.
Assume that the circumferential velocity (Vθ ) is linear across the gap at the each radius r.
ω
4 mm
D
Solution 6.81
At any radius r, the velocity Vθ is
 y
Vθ = rω   .
h
y
ω
4 mm
The shear stress at any radius r is
 dV 
 rω 
= μ .
τ = μ θ 
 h 
 dy  y=h
The torque due to the shear stress at radius r is
 rω 
dJ = (τ dA)r = r  μ
 (2π rdr )
 h 
2πμω R 3
πμωR 4
r dr =
Then J =
2h
h 0
N ⋅s
Figure A.3 gives μ = 0.088 2
m
so
J=


π  0.088
N ⋅s 
rev   rad 
4
1800
  2π
 (0.5 m)
2 
min  
rev 
m 
 sec 
2(0.004 m)  60

 min 
or
J = 410 N ⋅ m
r
D
Problem 6.82
lb
A viscous fluid (specific weight = 80
; viscosity = 0.03
lb ⋅ s
) is contained between two
ft
ft 2
infinite, horizontal parallel plates as shown in the figure below. The fluid moves between
the plates under the action of a pressure gradient, and the upper plate moves with a velocity
U while the bottom plate is fixed. A U-tube manometer
3
U = 0.02 ft/s
1.0 in.
y
6 in.
x
Fixed
plate
0.1 in.
γ = 100 lb/ft3
connected between two points along the bottom indicates a differential reading of 0.1 in. If
ft
the upper plate moves with a velocity of 0.02 , at what distance from the bottom plate
s
does the maximum velocity in the gap between the two plates occur? Assume laminar flow.
Solution 6.82
u =U
y 1  ∂p  2
+
  ( y − by )
b 2 μ  ∂x 
ℓ
(1)
Maximum velocity will occur at distance ym where
Thus,
du
= 0.
dy
1  ∂p 
du U
= +
  (2 y − b )
dy b 2 μ  ∂x 
And for
ym = −
γgf
du
=0
dy
μU
 ∂p 
b 
 ∂x 
+
b
2
(1)
For manometer (see the figure),
p1 + γ f Δh − γ gf Δh = p2
Or




lb
0.1in.
lb


= 0.167 2
p1 − p2 = (γ gf − γ f )Δh = 100 3 − 80 3  

ft
ft   12 in. 
ft

ft 

lb
γf
(2)
Δh
lb
0.167 2
∂p p1 − p2
ft = 0.334 lb
=
=
Also, −
∂x


ft3
 6 in. 
 in. 
 12

 ft 
thus, from Eq. (1)
1.0 in.
lb ⋅ s  
ft 

in.
 0.03 2   0.02 s  12
ft  
 +
ft = 0.0632 ft 12 in.  = 0.759 in.
ym = − 


2


 ft 
 1.0 in.  
lb 
 in.   −0.334 3 
ft 
 12

ft 

Problem 6.83
A flat block is pulled along a horizontal flat surface by a horizontal rope perpendicular to
one of the sides. The block measures 1.0 m ×1.0 m, has a mass of 100 kg and a constant vem
locity of 1.0 , and is separated from the flat surface by a 0.1-cm-thick oil layer of 15 °C
s
SAE 20 crankcase oil. Find the coefficient of sliding friction for the block. Does this coefficient of friction change with the velocity?
Solution 6.83
The force resisting the block’s motion is F = τ A.
Since the gap is thin, assume that the velocity profits in the oil is linear so
τ =μ
U
h
With U as the block velocity and h as the gap thickness. Then
μUA
.
h
The coefficient of friction ( μF ) is the ratio of the resisting force to the normal force, in this
F=
case, the block’s weight so
μF =
F μUA
=
W mgh
Figure A.1 gives μ = 0.300
(1)
N ⋅s
m2
, so
N ⋅ s  m 

1  (1 m 2 )
 0.300
2 
m  s 
μF = 
m

(100 kg)  9.81 2  (0.001 m)
s 

 s2 ⋅ N 


 kg ⋅ m 
μ F = 0.306, Eq. (1) shows μF is a function of velocity.
Problem 6.84
A viscosity motor/pump is shown in the figure below. The rotor is concentric within a stationary housing. The clearance h between the housing and the rotor is small compared to
the width w and radius R of the rotor. A seal divides the clearance space as shown.When
the device is operated as a motor, the viscous liquid is introduced at the high-pressure side
of the seal and leaves at the low-pressureside of the seal. This flow causes the rotor to turn
and produce power. For R = 1.0 ft, h = 0.50 in., w = 1.0 ft , ω = 10 rpm , and a flowrate of
ft3
1.5
, find the power output of the rotor.The fluid is 60 °F, SAE 30 crankcase oil. Nemin
glect gravity.
h
Housing
Rotor
+
R
ω
phigh
plow
Seal (no drag on rotor)
Solution 6.84
u = Rω
y
q
x
Since h << R, we can neglect the curvature and model the flow as that between parallel
 ∂(velocity ) 
plates with fully developed flow 
 = 0. We will assume the two plates are infi∂x


 ∂( anything ) 
nitely wide so there is no variations in the z direction 
 = 0. We will also as∂z


sume p = p ( x ) and p ≠ p( y, z ). Then for fully developed flow,
∂p dp plow − phigh
=
=
∂x dx
2π R
The flow is a combination of flow with a pressure gradient and flow with one moving plate.
From a force balance on a fluid particle, the velocity must satisfy
d 2u
dy
2
=
1 ∂p
.
μ ∂x
Integrating twice gives
u=
1 dp 2
y + C1y + C2
2 μ dx
The boundary conditions are
u = 0 at y = 0 and u = U at y = h.
Solving for C1 and C 2, the velocity profile gives
u =U
y 1 dp 2
( y − hy )
+
h 2 μ dx
where
dp
dp
> 0 is a pump and
< 0 is a motor.
dx
dx
(rotor drives fluid)
(fluid drives rotor)
The flowrate is
h
h y
1 dp 2

Q =  uwdy = w  U +
( y − hy )  dy
0
0  h 2u dx

h
Uy2 1 dp  y3 hy2   Uwh wh3 dp
=w 
+
−
 −
 =
2  
2
12 μ dx
 2 h 2 μ dx  3
0
(1)
The torque on the rotor is due to shear stress
J =  τ dA = 
2π 
0
 du 
du 
.
 μ  wRdθ = 2π Rwμ  
 dy h
 dy  y=h
Using the velocity profile gives
du U 1 dp
 U h dp 
(2 y − h ) and J = 2πμ Rw  +
= +
.
dy h 2 μ dx
 h 2u dx 
The power transferred to the rotor is
U
h dp  2πμU 2 w
dp
+ πUwh
W = J ω = 2πμ ( Rω )w  +
=
h
dx
 h 2 μ dx 
(2)
For the given motor problem,
 rad 
 2π

ft
rev 
 rev  
= 1.05
U = Rω = (1ft) 10

s
 min   60 s 


 min 
Figure A.2 gives
2
lbm

−6 lb ⋅ s   144 in   32.2 ft ⋅ lbm 
μ =  25 ×10

 = 0.116 ft ⋅ s
2 
2 
2


in   1ft   lb ⋅ s


The pressure gradient is found from Eq. (1) to be
dp 12 μ  Uwh

= 3
−Q 
dx wh  2

(3)
Substituting Eq. (3) into Eq. (2) gives
W=
2
2
2πμU 2w
 12 μ  Uwh
 2πμU w 6πμU w 12πμUQ
+ πUwh  3  
−Q  =
+
−
h
h
h
h2

 wh   2
8πμU 2w 12πμUQ 4πμU 
3Q 
=
−
=
 2Uw −

2
h
h 
h 
h
The numerical values give


ft3   min  
lbm 
ft  

3
1.5

4π  0.116
1.05 
 min   60 s  
ft
ft ⋅ s 
s 



W=
2(1.05
)(1 ft)- 

sec
 1ft 
 1ft  

0.50 in 
(0.50
in)


 

 12 in 
 12 in  
2
2

 lbm   ft  lb ⋅ s
= 11.0 




 ft ⋅ s2   s 
 32.2 ft ⋅ lbm 
W = 0.342
ft ⋅ lb
.
sec
Problem 6.85
A vertical shaft passes through a bearing and is lubricated with an oil having a viscosity of
N ⋅s
0.2 2 as shown in the figure below. Assume that the flow characteristics in the gap
m
between the shaft and bearing are the same as those for laminar flow between infinite
parallel plates with zero pressure gradient in the direction of flow. Estimate the torque rerev
quired to overcome viscous resistance when the shaft is turning at 80
.
min
Shaft
75 mm
Bearing
160 mm
0.25 mm
Oil
Solution 6.85
The torque due to force dF acting on a differential area, dA = ri dθ , is (see in the figure below)
dθ
ri
dF = τ dA
= τ ri ℓdθ
ℓ ~ shaft length
dJ = ri dF = ri 2τ dθ
where τ is the shearing stress. Thus,
J = ri 2τ l 
2π
0
dθ = 2π ri 2τ l
(1)
In the gap,
u =U
y
b
where U = ri ω and b is the gap width. Also,
τ =μ
du μU
=
dy
b
Thus, from Eq. (1)
U
l
N ⋅ s   rev  2π rad   min   (0.160 m)
 0.075  
J = 2π ri  μ  l = 2π ri 3 μω = 2π 
m   0.2 2   80



b
 b
 2
 
m   min  rev   60s   (0.25 ×10−3 m)
2
= 0.355 N ⋅ m
3
Problem 6.86
A viscous fluid is contained between two long concentric cylinders. The geometry of the
system is such that the flow between the cylinders is approximately the same as the laminar
flow between two infinite parallel plates. (a) Determine an expression for the torque required to rotate the outer cylinder with an angular velocity ω . The inner cylinder is fixed.
Express your answer in terms of the geometry of the system, the viscosity of the fluid,
and the angular velocity. (b) For a small, rectangular element located at the fixed wall,
determine an expression for the rate of angular deformation of this element. (See the
figure below.)
C
C'
τ xz
τ xz
σxx
τ xy
y
τ xy
D
D'
σxx
B
B'
A
A'
(b)
(a)
x
z
Solution 6.86
U = r0 ω
y
r0 – ri = b
u
τ
x
ω
τ
dθ
ri
r0
ℓ ~ cylinder length
τ ~ shearing stress
(a) The torque which must be applied to outer cylinder to overcome the force due to the
shearing stress is (see the figure above)
dJ = r0dF = r0 (τ r0 dθ ) = r02τ dθ
So that
J = r02τ 
2π
0
dθ = 2π r02τ
(1)
In the gap
y
u =U
b
Since, τ = μ
du μU
=
dy
b
and b = r0 − ri , U = r0ω (see the figure above), it follows from Eq. (1) that
 μr ω 
2π r03 μω
J = 2π r02  0  =
r0 − ri
 r0 − ri 
(b) From the equation
γ=
∂v ∂u
+
∂x ∂y
For the linear distribution
u=−
r0ω
Uy
y=−
r0 − ri
b
So that
U
∂u
=−
∂y
b
And since v = 0
γ =−
U
b
The negative sign indicates that the original right angle show in Figure (b) in the problem is
increasing.
Problem 6.87
Verify that the momentum correction factor β for fully developed, laminar flow in a circu4
lar tube is .
3
Solution 6.87
The fully developed velocity profile for the axial velocity u is

r2 
u = umax  1 − 2  .
 R 


Equation (7.41) gives
A u dA N
β=
=
2
V 2A
D
where V is the average velocity defined by
V=
1
R
2umax
 u2π rdr = R2
π R2 0
R
R
2umax  r 2
2umax  R2 R2  umax
r2 
r4 
1
rdr
−
=
−
=
−
 =
0  R2 
2 2
2
2 
 2
4
2
R
4
R
R





0


Now
2
R 2
R
R
r2 
2r 2 r 4 
2
u 2π rdr = 2π u max 1 − 2  rdr = 2π u2 max 1 − 2 + 4  rdr
0
0 
0 
R 
R
R 
N =


R




 r2
 R2 R2 R6  π R2 u2 max
r4
r6 
= 2π u max  − 2 + 4  = 2π u2 max 
−
+
 =
 2
2
6
3
 2 2R 6R  0


2
Then
 π R2  2

 u max
N  3 
4
= β= .
β= =
2
D  umax 
3
2
 2  πR


Problem 6.88
Verify that the kinetic energy correction factor α for fully developed, laminar flow in a circular tube is 2.0.
Solution 6.88
The fully developed velocity profits for the axial velocity u is

r2 
u = umax  1 − 2  .
 R 


Equation (7.42) gives
A u dA N
α=
=
3
V 3A
D
where V is the average velocity defined by
V=
1
R
2umax
 u2π rdr = R2
π R2 0
R
R
2umax  r 2
2umax  R2 R2  umax
r2 
r4 
−
=
−
=
−
1
rdr


 =


0  R2 
2
2
2 
 2
2
4
2
R
4
R
R





0


Now
3
R 3
R
R
r2 
3r 2 3r 4 r 6 
3
−
u 2π rdr = 2π u max 1 − 2  rdr = 2π u3max 1 − 2 +
 rdr
0
0 
0 
R 
R
4R 4 R 6 
N =



R



 r2
 R2 3R2 R2 R2  π R2u3max
3r 4 3r6
r8 
= 2π u max  − − 2 + 7 − 6  = 2π u3max 
−
+
−
=
 2
2
2
8 
3
4R 6R 8R  0
 2

3
Then
 π R3  3

 u max
N  4 
= α = 2.0 .
α= =
D  umax 3
2
 2  πR


Problem 6.89
A simple flow system to be used for steady-flow tests consists of a constant head tank connected to a length of 4-mm-diameter tubing as shown in the figure below. The liquid has a
N ⋅s
kg
, a density of 1200 3 , and discharges into the atmosphere with a
viscosity of 0.015
2
m
m
m
mean velocity of 2 . (a) Verify that the flow will be laminar. (b) The flow is fully devels
oped in the last 3 m of the tube. What is the pressure at the pressure gage? (c) What is the
magnitude of the wall shearing stress, τ rz , in the fully developed region?
Pressure
gage
Diameter = 4 mm
3m
Solution 6.89
(a) Check Reynolds number to determine if flow is laminar (See Chapters 7 and 8):
kg  m 

1200 3  2  (0.004 m)

ρV (2R ) 
m  s 
=
= 640
Re =
N ⋅s
μ
0.015 2
m
Since the Reynolds number is well below 2100, the flow is laminar.
(b) For laminar flow,
R 2 Δp
V=
8μ l
Since Δp = p1 − p2 = p1 − 0 (see the figure in the Problem)
N ⋅ s  m 

8  0.015 2  2  (3 m)
8μV
m  s 
= 
= 180 kPa
p1 =
2
2
R
 0.004 m 


2


 ∂v ∂v 
(c) τ rz = μ  r + z 
∂r 
 ∂z
For fully developed pipe flow, vr = 0 , so that
τ rz = μ
∂vz
∂r
  r 2 
Also, vz = vmax 1 −   
  R  
And with vmax = 2V , where V is the mean velocity
 2r 
τ rz = 2V μ  − 2 
 R 
Thus, at the wall, r = R,
N ⋅s 
 m 
4  2   0.015 2 
−4V μ
s 
m  = 60.0 N
(τ rz )wall =
=− 
R
 0.004 
m2
m

 2

Problem 6.90
(a) Show that for Poiseuille flow in a tube of radius R , the magnitude of the wall shearing
stress, τ rz , can be obtained from the relationship
4 μQ
(τ rz )wall =
π R3
for a Newtonian fluid of viscosity μ. The volume rate of flow is Q . (b) Determine the magN ⋅s
flowing with an
nitude of the wall shearing stress for a fluid having a viscosity of 0.004
m2
mm
in a 2-mm-diameter tube.
average velocity of 130
s
Solution 6.90
 ∂v ∂v 
(a) τ rz = μ  r + z 
∂r 
 ∂z
For Poiseuille flow in a tube, vr = 0, and therefore
τ rz = μ
∂vz
∂r
  r 2 
Since, vz = vmax 1 −   
  R  
And vmax = 2V , where V is the mean velocity, it follows that
∂v z
4V
= − 2r
∂r
R
Thus, at the wall (r = R),
(τ rz )wall = −
4 μV
R
and with Q = π R 2V
(τ rz )wall =
4 μQ
π R3
N ⋅s 
m

4  0.004 2   0.130 
4 μV
s 
m 
(b) (τ rz )wall =
= 
= 2.08Pa
R
 0.002 
m


 2

Problem 6.91
An infinitely long, solid, vertical cylinder of radius R is located in an infinite mass of an incompressible fluid. Start with the Navier–Stokes equation in the θ direction and derive an
expression for the velocity distribution for the steady-flow case in which the cylinder is rotating about a fixed axis with a constant angular velocity ω . You need not consider body
forces. Assume that the flow is axisymmetric and the fluid is at rest at infinity.
Solution 6.91
For this flow field, vr = 0, vz = 0, and from the continuity equation,
1 ∂( rvr ) 1 ∂vθ ∂vz
+
+
=0
∂z
r ∂r
r ∂θ
It follows that
∂vθ
= 0 (see the figure below)
∂θ
vr = 0
vθ
r
θ
ω
R
Thus, the Navier–Stokes equation in the θ -direction for steady flow reduces to
0=−
 1 ∂  ∂vθ  vθ 
1 ∂p
+μ
r
− 2 
r ∂θ
 r ∂r  ∂r  r 
Due to the symmetry of the flow,
∂p
=0
∂θ
So that
1 ∂  ∂vθ  vθ
− =0
r
r ∂r  ∂r  r 2
Or
∂ 2vθ
+
1 ∂vθ vθ
− =0
r ∂r r 2
(1)
∂r
Since vθ is a function of only r, Eq. (1) can be expressed as an ordinary differential equation, and rewritten as
2
d 2vθ
dr
+
2
d  vθ 
 =0
dr  r 
(2)
Equation (2) can be integrated to yield
dvθ vθ
+ = C1
dr
r
Or
r
dvθ
+ vθ = C1r
dr
(3)
Equation (3) can be expressed as
d ( rvθ )
= C1r
dr
And a second integration yields
rvθ =
C1r 2
+ C2
2
Or
C1r C2
+
2
r
vθ =
As r → ∞ , vθ → 0 ( since fluid is at rest at infinity)
So that C1 = 0 . Thus,
vθ =
C2
r
and since at r = R, vθ = Rω , it follows that C2 = R2ω
And
vθ =
R2ω
r
Problem 6.92
We will see in Chapter 8 that the pressure drop in fully developed pipe flow is sometimes
computed with the aid of a friction factor, defined by
Δp D
f =
1
ρV 2
2
where V is the average velocity, D is the pipe diameter, and is the length of pipe over
which Δp occurs. For laminar fully developed flow, f can be evaluated from
C
f =
Re
where C is a constant and Re is the Reynolds number, given by Re = ρVD / μ . Find the constant C for flow in a circular tube.
Solution 6.92
See Section 8.2 in the text where it is shown that C = 64.
Problem 6.93
A liquid (viscosity = 0.002
N ⋅s
2m
; density = 1000
kg
) is forced through the circular tube
m3
m
shown in the figure below. A differential manometer is connected to the tube as shown to
measure the pressure drop along the tube. When the differential reading, Δh, is 9 mm, what
is the mean velocity in the tube?
2
4 mm
Δh
Density of
gage fluid = 2000 kg/m3
(1)
(2)
γf
For manometer (see the figure),
p1 + γΔh − γ gf Δh = p2
4 mm
2m
Solution 6.93
Assume laminar flow so that
R 2 Δp
V=
8μ
Δh
Density of
gage fluid = 2000 kg/m3
γgf
Or
m 
kg
kg 
N

p1 − p2 = Δp = Δh(γ gf − γ ) = Δh( g )( ρ gf − ρ f ) = (0.009 m)  9.81 2   2000 3 − 1000 3  = 88.3 2
s 
m
m 
m

2
N 
 0.004  
m   88.3 2 

3
m  = 1.10 ×10 −2 m
 
Thus, V = 
N ⋅s 
s

8  0.002 2  (2 m)
m 

Check Reynolds number to confirm that flow is laminar:
m
 3 kg 
10 3 1.10 ×10−2  (0.004 m)

ρV (2R ) 
s 
m 
Re =
=
= 22.0 < 2100
N ⋅s
μ
0.002 2
m
Since Re < 2100 flow is laminar.
Problem 6.94
Fluid with kinematic viscosity v flows down an inclined circular pipe of length and
diameter D with flowrate Q. Find the vertical drop per unit length of the pipe so that the
pressure drop ( p1 − p2 ) is zero for laminar flow.
Solution 6.94
The mechanical energy equation is written from 1 to 2 to get
p1
ρ
+
α1V12
2
+ gz1 =
p2
ρ
+
α 2V22
2
+ gz2 + gh
Assuming fully developed flow and constant density, α1V12 = α2V22
Problem statement gives p1 − p2 = 0 . Then
z1 − z2 =
gh
g
For laminar flow,
gh =
128v Q
π D4
So
z1 − z2 =
128v Q
π gD 4
Problem 6.95
Blood flows at volume rate Q in a circular tube of radius R . The blood cells concentrate and
flow near the center of the tube, while the cell-free fluid (plasma) flows in the outer region.
The center core of radius Rc has a viscosity μ c , and the plasma has a viscosity μp . Assume
laminar, fully developed flow for both the core and plasma flows and show that an
“apparent” viscosity is defined by μapp =
is given by μapp =
π R 4 Δp
8LQ
μp
4
 Rc   μp 
1− 

 1 −
 R   μc 
Solution 6.95
We begin with a force balance on a cylindrical fluid element, this gives
μ
du
rδp
=−
dr
2 δx
where
μ = μc r < Rc , μ = μ p Rc < r < R ,
And
δ p Δp
=
= constant .
δx L
The velocity distribution in both fluids is found by integrating
μc = −
1 δp 2
1 δp 2
r + C1 and μ p = −
r + C2
4 μc δ x
4 μc δ x
The constants are evaluated from the conditions
μp = 0
at r = R
(A)
at r = Rc
(B)
and
μ p = μc
From (A)
0=−
1 δp 2
R + C2
4μ p δ x
so
C2 =
1 δp 2
R
4μ p δ x
From (B)
−
1 δp 2
1 δp 2
Rc + C1 = −
( Rc − R 2 )
4 μc δ x
4μ p δ x
so
C1 =

1 δ p  μ p  2
− 1 Rc + R 2 .

4 μ p δ x  μc


The flowrate is calculated by
Rc
R
0
Rc
Q =  μc 2π rdr +  μ p 2π rdr,


Rc 
R 
Q
1 δp 3
1 δp 3
r + C1r  dr +   −
=   −
 r + C2 r  dr

0
Rc  4 μ δ x 
2π


p
 4 μc δ x 


R
R
 1 δ p  r 4 C1r 2  c  1 δ p  r 4 C2 r 2 
=  −

 +
 +  −
 +
 4
2 
4
μ
δ
2
x

 R
p
 4 μc δ x  4


0
c
 1 δ p  Rc4 C1Rc2  1 δ p  R 4 − Rc4 C2 ( R 2 − Rc2 )
= −
+
+−
+
.


 4μ p δ x 
2
4
2
 4 μc δ x  4


Substituting C 1 and C 2 gives
2 


Q
1 δ p  μ p Rc4 Rc4  μ p   R     1 δ p   R 4 − Rc4 1 2 2
 1 −
=−
+
+ ( Rc R − R 4 )

 −    − 
2π
4 μ p δ x  μc 4
2  μc   Rc     4 μ p δ x  
4
2




and
4
4
4
2
4
2

R 4 δ p  μ p  Rc 
 Rc  μ p
 Rc 
 Rc   Rc 
 Rc 
=
+
2
−
2
+
2
+
−
1
−
2
+
2
−

R μ
R
R R
R
16 μ p δ x  μc  R 
  c
 
   
 

Noting
8QL
πΔpR
δ p Δp
= , we can collect terms to get
δx L
=
4
4
4
1   Rc  μ p  Rc  
1 −   +

μ p   R  μc  R  
Inverting both sides gives
μapp =
πΔpR 4
8QL
=
μp
4
 R   μp 
1 −  c  1 −

 R   μc 
Problem 6.96
An incompressible Newtonian fluid flows steadily between two infinitely long, concentric
cylinders as shown in the figure below. The outer cylinder is fixed, but the inner cylinder
moves with a longitudinal velocity V0 as shown. The pressure gradient in the axial direction
−Δp
. For what value of V0 will the drag on the inner cylinder be zero? Assume that the
is
flow is laminar, axisymmetric, and fully developed.
Fixed wall
ro
V0
ri
Solution 6.96
The equation below for flow in circular tubes applies in the annular region. Thus,
vz =
1  ∂p  2
  r + C1 ln r + C2
4 μ  ∂z 
(1)
Fixed wall
r
vz
Z
ro
V0
ri
With boundary conditions, r = r0, vz = 0, and r = ri, vz = V0,
it follows that:
0=
1  ∂p  2
  r0 + C1 ln r0 + C2
4 μ  ∂z 
(2)
1  ∂p  2
  ri + C1 ln ri + C2
4 μ  ∂z 
(3)
V0 =
Subtract Eq. (2) from Eq. (3) to obtain
V0 =
ri
1  ∂p  2
2
  ( ri − r0 ) + C1 ln
ro
4 μ  ∂z 
So that C1 =
1  ∂p  2
2
  ( ri − r0 )
4 μ  ∂z 
r
ln i
r0
V0 −
The drag on the inner cylinder will be zero if
(τ rz )r =ri = 0
 ∂v ∂v 
Since, τ rz = μ  r + z 
 ∂z ∂r 
And with vr = 0, it follows that
τ rz = μ
∂vz
∂r
Differentiate Eq. (1) with respect to r to obtain
∂
∂ 
=
+
∂
μ  ∂ 
So that at r = ri

1  ∂p  2
2 
V0 −
  ( ri − r0 ) 
 1  ∂p 
4 μ  ∂z 

(τ rz )r =ri = μ    ri +
ri
 2 μ  ∂z 

ri ln


r0
Thus, in order for the drag to be zero,
1  ∂p 
  ri +
2 μ  ∂z 
V0 −
1  ∂p  2
2
  ( ri − r0 )
4 μ  ∂z 
=0
ri
ri ln
r0
Or
V0 = −
1  ∂p   2 ri
2
2 
  2ri ln − ( ri − r0 )
4 μ  ∂z  
r0

Problem 6.97
A viscous fluid is contained between two infinitely long, vertical, concentric cylinders. The
outer cylinder has a radius ro and rotates with an angular velocity ω .The inner cylinder is
fixed and has a radius ri . Make use of the Navier–Stokes equations to obtain an exact
solution for the velocity distribution in the gap. Assume that the flow in the gap is axisymmetric (neither velocity nor pressure are functions of angular position θ within the gap) and
that there are no velocity components other than the tangential component. The only body
force is the weight.
Solution 6.97
ω
vθ
vr = 0
r
ri
r0
The velocity distribution in the annular space is given by the equation
Cr C
(1)
vθ = 1 + 2
2
r
With the boundary conditions r = ri , vθ = 0 , and r = r0, vθ = r0ω (see the figure above), it follows
From Eq. (1) that :
Cr C
0= 1i + 2
ri
2
Cr C
r0ω = 1 0 + 2
2
r0
Therefore ,
2ω
C1 =
ri 2
1− 2
r0
And
C2 =
−ri 2ω
r2
1− i 2
r0
So that
v0 =
rω
1−
ri 2
r0 2
−
ri 2ω
 ri 2 
r 1 − 2 
 r 
0 

Or
rω  ri 2 
v0 =
1 − 
ri 2  r 2 
1− 2
r0
Problem 6.98
A double-pipe heat exchanger consists of two concentric tubes with one fluid flowing in the
central tube and the other flowing in the annulus between the tubes. In a particular
exchanger, cold (15 C) water flows at 300 l/min through a 10-cm diameter inner tube and
hot (50 C) SAE 30 oil flows at 400 l/min in the annulus between the inner tube and a
12-cm diameter outer tube. The heat exchanger is 2 m in length. (a) Calculate the pressure
drop in the oil stream. (b) Can you calculate the pressure drop in the water using the same
equations? Why or why not?
Solution 6.98
Q=
πΔp  4 4 ( r02 − ri2 )2 
r0 − ri −

8μ 
ln( r0 ri ) 
8μ Q  4 4 ( r02 − ri2 )2 
Δp =
r − r −

ln( r0 / ri ) 
π  0 i
−1
For the oil: From Table 1.6, ρ ≈ 910 kg/m3 and from Figure B.2, ν ≈ 2.1×10−5 m2 /s
μ =ρν =0.019kg ⋅ m2 /s
8 × 0.019kg ⋅ m 2 /s × 2 m × 400
Δp =
π
l
1min/ 60s
×
2 2
2 2 2  −1
min 1000 l/m3 
4 4
4 4 (0.06 m − 0.05 m )
0.06 m − 0.05 m −

ln(0.06 m / 0.05 m) 

Δp = 8790 Pa = 8.79kPa
No. The water pressure drop cannot be calculated from the same equations because the water flows in a circular tube, not an annulus.
Problem 6.99
A wire of diameter d is stretched along the centerline of a pipe of diameter D. For a given
pressure drop per unit length of pipe, by how much does the presence of the wire reduce the
d
d
flowrate if (a) = 0.1; (b) = 0.01?
D
D
Solution 6.99
The volume flowrate is given by the equation



2
2 2
πΔp  4 4 ( r0 − ri ) 
Q=
r0 − ri −
8μ 
r  
ln  0  

 ri  

which can be written as
2

  r 2  

1 −  i   


4
π r0 4 Δp   ri    r0   
Q=
1 −   +

8μ   r0 
 ri  
ln  

 r0  




ri d
Since = , Eq. (1) can also be written as
r0 D
2

  d 2  

1 −    
4
4

π r Δp   d    D   
Q= 0
1 −   +

8μ   D 
d
ln   

D 


Note that for
Q=
(1)
(2)
d
= 0 (no wire)
D
π r0 4 Δp
8μ
which corresponds to Poiseuille’s Law.
(a) For
d
= 0.1, Eq. (2) gives
D
2 2
π r0 4 Δp 
4 [1 − (0.1) ] 
Q=
1 − (0.1) +
 = 0.574
8μ 
ln(0.1) 
Thus, for the same Δp, the flowrate is reduced by % reduction in
Q = (1 − 0.574) ×100 = 42.6%
(b) Similarly, for
Q=
d
= 0.01, Eq. (2) gives
D
2 2
π r0 4 Δp 
4 [1 − (0.01) ] 
−
+
1
(0.01)

 = 0.783
8μ 
ln(0.01) 
and % reduction in Q = (1 − 0.783) ×100 = 21.7%
Note that the presence of even a very small wire along the tube centerline has a significant
effect on the flowrate.
Problem 6.100
Look through several professional society magazines such as Civil Engineering, Mechanical
Engineering, or ASEE Prism. Count the advertisements for CFD software and calculate an
“ad density” in advertisements per issue.
Solution 6.100
This is an open-ended student activity for which student responses will vary. No Solution is
available.
Problem 6.101
Consider the Navier–Stokes equations as written in Eq. (6.158). Write the physical meaning
of each term in the equation; for example, “The term ρ g is the gravity force (or weight) per
unit volume.”
Solution 6.101
This is an open-ended student activity for which student responses will vary. No Solution is
available.
Problem 6.102
Find out what CFD software packages your college or university has. How many “seats” of
each? How much does your institution spend annually on CFD software?
Solution 6.102
This is an open-ended student activity for which student responses will vary. No Solution is
available.
Problem 7.1
A mixing basin in a sewage filtration plant is stirred by mechanical agitation (paddlewheel)
with a power input W (ft ⋅ lb/s). The degree of mixing of fluid particles is measured by a
“velocity gradient” G given by
G=
W
μV
where μ is the fluid viscosity in lb ⋅ s/ft 2 and V is the basin volume in ft3. Find the units of
the velocity gradient.
Solution 7.1
G=
W
.
μV
Substituting the units of W , μ , and V into the equation gives
 ft ⋅ lb 


1 1
[G ] = lb ⋅ ss  = 2 =
s

 3
s
 2  ft
 ft 
( )
or
[G ] =
1
s
Since G is termed a velocity gradient, it physically means a velocity change divided by a
distance or a frequency. Rearranging gives
[G ] =
ft/s
1/ ft
Problem 7.2
An equation used to evaluate vacuum filtration is
Q=
ΔpA2
α ( VRw + AR f )
,
where Q  L3 / T is the filtrate volume flowrate, Δp  F / L2 is the vacuum pressure
differential, A  L2 is the filter area, α is the filtrate “viscosity,” V  L3 is the filtrate
volume, R  L / F is the sludge specific resistance, w  F / L3 is the weight of dry sludge per
unit volume of filtrate, and Rf is the specific resistance of the filter medium. What are the
dimensions of Rf and α ?
Solution 7.2
Q=
ΔpA2
μ ( VRw + AR f )
Substituting the given dimensions into the equation gives
( )
F  2 2
 2 L
L3
L 

T
 3L F

+ L2 R f  
[μ ] L
3
 F L

=
FL2
[ μ ] ( L + L2 R f  )
The dimensions of R f should be such that
R f  L2  L or
 
Rf 
1
L
The dimensions of μ should be such that
μ
FL2
 L3 
  L
T 
or
μ
FT
L2
Problem 7.3
Verify the left-hand side of
DΔp
ρV
2
 ρVD 
=φ
 is dimensionless using the MLT system.
 μ 
Solution 7.3
 ρVD 
L
M
M
φ
=
,
where
D

L
,
,
,
and

V
p
Δ

ρ




T
ρV 2
L2T 2
L3
 μ 
DΔp
Thus,
M
DΔp
L2T 2 = M 0 L0T 0

 M L2 
ρV 2
 3 2 
L T 
L
That is
DΔp
ρV 2
is dimensionless.
Problem 7.4
The Reynolds number, ρVD / μ , is a very important parameter in fluid mechanics. Verify
that the Reynolds number is dimensionless, using both the FLT system and the MLT
system for basic dimensions, and determine its value for ethyl alcohol flowing at a velocity
of 3 m/s through a 2-in.- diameter pipe.
Solution 7.4
( −4 2 )(
ρVD FL T LT
Reynolds number =
=
μ
FL−2T
−1
) ( L ) = F 0L0T 0
ML−3 )( LT −1 ) ( L )
(
=
= M 0 L0T 0
ML−1T −1
For ethyl alcohol, μ = 1.19 × 10 −3
N ⋅S
kg
m
m3
and ρ = 789
3
.
kg   m  2  
m

789 3   3  ft   0.3048 

ρVD 
ft 
m   s  12  
Thus,
=
= 1.01× 105
⋅
N
S
μ
1.19 ×10 −3 2
m
Problem 7.5
For the flow of a thin film of a liquid with a depth h and a free surface, two important
dimensionless parameters are the Froude number, V / gh , and the Weber number,
ρV 2 h / σ . Determine the value of these two parameters for glycerin (at 20  C) flowing with a
velocity of 0.7 m/s at a depth of 3 mm.
Solution 7.5
0.7
m
s
V
=
= 4.08
gh
m


 9.81 2  ( 0.003 m )
s 

2
kg 
m

1260 3  0.7  ( 0.003 m )

2
ρV h 
s 
m 
=
= 29.3
σ
−2 N
6.33 ×10
m
Problem 7.6
The Mach number for a body moving through a fluid with velocity V is defined as V c,
where c is the speed of sound in the fluid. This dimensionless parameter is usually
considered to be important in fluid dynamics problems when its value exceeds 0.3. What
would be the velocity of a body at a Mach number of 0.3 if the fluid is (a) air at standard
atmospheric pressure and 20 °C, and (b) water at the same temperature and pressure?
Solution 7.6
V
(a) = 0.3
c
m
For air at 20 C, c = 343.3
s
so that
m
m

V = 0.3  343.3  = 103
s 
s

(b) For water at 20 C, c = 1481
m
s
so that
m
m

V = 0.3 1481  = 444
s 
s

Problem 7.7
A mixing basin in a sewage filtration plant is stirred by a mechanical agitator with a power
input W = F ⋅ L / T . Other parameters describing the performance of the mixing process are
the fluid absolute viscosity μ = F ⋅ T / L2, the basin volume V = L3, and the velocity gradient
G = 1/ T . Determine the form of the dimensionless relationship.
Solution 7.7
We will use the Pi Theorem to find the dimensionless parameter(s).
The dimensional parameters are:
FL
W 
T
FT
μ 2
L
V  L3
G
1
T
The number of fundamental dimensions is 3 ( F , L,T ). Therefore, k ≤ 3. Since μ , V , and
G cannot form a dimensionless group k = 3. n = number of dimensionless parameters
= n − k =1
Select μ ,V , and G as repeating parameters. Combining them with W gives
W  [ μ ] [V ] [G ] = F 0 L0T 0
a
b
c
or
a
c
 FL  FT 
b 1 
0 0 0

 2  ( L )   = F L T
 T  L 
T 
Equating power of like dimensions gives
a = −1

( 2a − 1) = −1

L : 1 − 2a + 3b = 0  b =
3
T : −1 + a − c = 0 
c = a − 1 = −2
F :1 + a = 0
The dimensionless parameter is
Π1 = W μ −1V −1G −2
or
Π1 =
W
μ VG 2
The form of the dimensionless relationship is
W
μ VG 2
= C , C a constant,
or
G
μV
W
= C ′ , C ′ a constant.
Problem 7.8
The excess pressure inside a bubble is known to be dependent on bubble radius and surface
tension. After finding the pi terms, determine the variation in excess pressure if we (a)
double the radius and (b) double the surface tension.
Solution 7.8
Δp = f ( R,σ ), where Δp =
F
2
L
=
M
LT
2
, R = L , and σ =
F M
=
L T2
Consider the ( MLT ) dimensions so that k − r = 3 − 3 = 0 since there are 3 variables and
3 dimensions.
According to this, there should be k − r = 0 pi terms!?
However, if we consider the ( FLT ) dimensions we see that it takes only F and L ; T is not
needed, so that r = 2 .
Hence, k − r = 3 − 2 = 1, so only 1 pi term is needed.
That is, Π1= constant
To determine Π1, consider Π1 = ΔpRaσ b or
ΔpRaσ b =
Thus:
F:
1+ b = 0
L:
a−b−2 = 0
or b = −1 and a = b + 2 = 1
Hence Π1 =
ΔpR
σ
or
ΔpR
σ
= C , where C = constant.
or
Δp =
Cσ
R
(1)
(a)
If R is doubled, Δp is reduced by half . (See Eq. (1))
(b)
If σ is doubled, Δp is doubled . (See Eq. (1))
b
aF 
1+ b −2+ a −b
L
  =F L
2
L
L
F
Problem 7.9
At a sudden contraction in a pipe, the diameter changes from D1 to D2. The pressure drop,
Δp, which develops across the contraction is a function of D1 and D2, as well as the velocity,
V , in the larger pipe, and the fluid density, ρ , and viscosity, μ . Use D1, V , and μ as repeating
variables to determine a suitable set of dimensionless parameters. Why would it be incorrect
to include the velocity in the smaller pipe as an additional variable?
Solution 7.9
Δp = FL−2
Δp = f ( D1, D2 ,V , ρ , μ )
D1 = L
ρ = FL−4T 2
V = LT −1
D2 = L
μ = FL−2T
From the pi theorem, 6 − 3 = 3 dimensionless parameters required.
Use D1, V , and μ as repeating variables. Thus,
Π1 = ΔpD1aV b μ c
and
( FL )−2 ( L )a ( LT −1 ) ( FL−2T ) = F 0 L0T 0
b
c
so that
1+ c = 0
(for F )
−2 + a + b − 2c = 0
(for L )
−b + c = 0
(for T )
a = 1, b = −1, c = −1, and therefore,
It follows that
Π1 =
ΔpD1
Vμ
Check dimensions using MLT system:
(
)
ML−1T −2 ( L )
ΔpD1
=
= M 0 L0T 0
−1
−1 −1
Vμ
LT
ML T
(
)(
)
For Π 2:
Π2 = D2D1aV b μ c
L (L)
a
( LT −1 ) ( FL−2T ) = F 0L0T 0
b
c
c=0
(for F )
1 + a + b − 2c = 0
(for L )
−b + c = 0
(for T )
It follows that a = −1, b = 0, c = 0, and therefore
Π2 =
Π 2 is obviously dimensionless.
D2
D1
∴ ok
For Π3:
Π3 = ρ D1aV b μ c
(
)
−2
( L )a ( LT −1 ) ( FL−2T ) = F 0 L0T 0
b
c
1+ c = 0
(for F )
−4 + a + b − 2c = 0
(for L )
2−b+c = 0
(for T )
FL−4T 2
a = 1, b = 1, c = −1, and therefore
It follows that
Π3 =
ρ DV
1
μ
Check dimensions using MLT system:
(
) (
ML ( L ) LT
ρ DV
1 
μ
ML−1T −1
−3
−1
) = M L T
0 0
0
∴ ok
Thus,
 D ρ DV 
ΔpD1
=φ 2 , 1 
Vμ
μ 
 D1
From the continuity equation,
V
π
4
D12 = Vs
π
4
D2 2
where Vs is the velocity in the smaller pipe. Since
2
D 
Vs =  1  V
 D2 
Vs is not independent of D1, D2, and V , and therefore, should not be included as an
independent variable.
Problem 7.10
Water sloshes back and forth in a tank as shown in the figure below. The frequency of
sloshing, ω , is assumed to be a function of the acceleration of gravity, g , the average depth
of the water, h , and the length of the tank, . Develop a suitable set of dimensionless
parameters for this problem using g and  as repeating variables.
ω
h
ℓ
Solution 7.10
ω = f ( g , h,  )
ω = T −1
h = L
 = L
g = LT −2
From the pi theorem, 4 − 2 = 2 dimensionless parameters required. Use g and  as repeating
variables, thus,
Π1 = ω g a b
and
(T −1 )( LT −2 ) ( L )b = L0T 0
a
so that
a+b =0
−1 − 2a = 0
It follows that
(for L )
(for T )
1
1
a = − , b = , and therefore
2
2

Π1 = ω
g
Check dimensions:
 1
=
g T
ω
L
LT
−2
= L0T 0
∴ ok
For Π 2:
Π2 = hg a b
(
L LT −2
) ( L )b = L0T 0
a
1+ a + b = 0
(for L )
−2a = 0
(for T )
a = 0, b = −1, and therefore
It follows that
h
Π2 =

and Π 2 is obviously dimensionless. Thus,
ω

h
=φ 
g

Problem 7.11
Assume that the flowrate, Q, of a gas from a smokestack is a function of the density of the
ambient air, ρa , the density of the gas, ρg , within the stack, the acceleration of gravity, g ,
and the height and diameter of the stack, h and d , respectively. Use ρa , d , and g as
repeating variables to develop a set of pi terms that could be used to describe this problem.
Solution 7.11
(
Q = f ρa , ρg , g , h, d
ρa  ML−3
Q  L3T −1
)
ρg  ML−3
g  LT −2
hL
d L
From the pi theorem, 6 − 3 = 3 pi terms required. Use ρa , d , and g as repeating variables.
Thus,
Π1 = Qρaa d b g c
and
(
)(
L3T −1 ML−3
) ( L )b ( LT −2 ) = M 0L0T 0
a
c
so that
a=0
3 − 3a + b + c = 0
−1 − 2c = 0
(for M )
(for L )
(for T )
5
1
a = 0, b = − , c = − , and therefore
2
2
Q
Π1 = 5 1
It follows that
d 2g2
Check dimensions using FLT system:
Q
5 1
d 2g2
=
L3T −1
1
5
−2 2
L 2 LT
( ) (
)
∴ ok
= F 0 L0T 0
For Π 2:
Π 2 = ρg ρaa d b g c
(ML− )(ML− ) ( L )b ( LT − ) = M L T
3
3 a
2 c
1+ a = 0
3 − 3a + b + c = 0
−2c = 0
It follows that
0 0
0
(for M )
(for L )
(for T )
a = −1, b = 0, c = 0, and therefore
Π2 =
which is obviously dimensionless.
ρg
ρa
For Π3:
Π3 = hρaad b gc
( L ) ( ML−3 ) ( L )b ( LT −2 ) = M 0 L0T 0
a
a=0
1 − 3a + b + c = 0
−2c = 0
c
(for M )
(for L )
(for T )
a = 0, b = −1, c = 0, and therefore
It follows that
Π3 =
which is obviously dimensionless.
Thus,
Q
5
d2g
 ρg h 
=
φ
 , 
1
 ρa d 
2
h
d
Problem 7.12
The pressure rise, Δp, across a pump can be expressed as
Δp = f ( D, ρ ,ω,Q )
where D is the impeller diameter, ρ is the fluid density, ω is the rotational speed, and Q is
the flowrate. Determine a suitable set of dimensionless parameters.
Solution 7.12
Δp = FL−2
ω  T −1
ρ = FL−4T 2
D = L
Q = L3T −1
From the pi theorem, 5 − 3 = 2 pi terms required. Use D, ρ , and ω as repeating variables.
Thus,
Π1 = ΔpDa ρ bωc
and
( FL− ) ( L )a ( FL− T ) (T − ) = F L T
2
2 b
4
1 c
0 0
0
so that
(for F )
1+ b = 0
(for L )
−2 + a − 4b = 0
(for T )
2b − c = 0
a = −2, b = −1, c = −2, and therefore
Δp
Π1 = 2 2
D ρω
It follows that
Check dimensions using MLT system:
Δp
D ρω
2
=
2
ML−1T −2
( L ) ( ML−3 )(T
2
)
−1 2
= M 0 L0T 0 ∴ ok
For Π 2:
Π2 = QDa ρ bω c
( L T − ) ( L )a ( FL− T ) (T − ) = F L T
3
1
4
2 b
1 c
b=0
3 + a − 4b = 0
−1 + 2b − c = 0
0 0
0
(for F )
(for L )
(for T )
a = −3, b = 0, c = −1, and therefore
Q
Π2 = 3
Dω
Check dimensions using MLT system:
It follows that
Q
Dω
3
=
L3T −1
( L ) (T
3
−1
)
= M 0 L0T 0
Thus,
Δp
 Q 
=φ 3 
D ρω
D ω
2
2
∴ ok
Problem 7.13
A thin elastic wire is placed between rigid supports. A fluid flows past the wire, and it is
desired to study the static deflection, δ , at the center of the wire due to the fluid drag.
Assume that
δ = f ( , d , ρ , μ,V , E )
where  is the wire length, d is the wire diameter, ρ is the fluid density, μ is the fluid
viscosity, V is the fluid velocity, and E is the modulus of elasticity of the wire material.
Develop a suitable set of pi terms for this problem.
Solution 7.13
δ = L
 = L
ρ = FL−4T 2
d = L
μ = FL−2T
V = LT −1
E = FL−2
From the pi theorem, 7 − 3 = 4 pi terms required. Use d , V , and E as repeating variables.
Thus,
Π1 = δ d aV b E c
and
( L )( L )a ( LT −1 ) ( FL−2 ) = F 0 L0T 0
b
c
so that
c=0
1 + a + b − 2c = 0
−b = 0
(for F )
(for L )
(for T )
a = −1, b = 0, c = 0, and therefore
δ
Π1 =
d
which is obviously dimensionless.
It follows that
For Π 2:
Π2 = d aV b E c
and as for Π1, a = −1, b = 0, c = 0, so that
Π2 =

d
For Π 3:
Π3 = ρ d aV b E c
( FL T ) ( L )a ( LT ) ( FL ) = F L T
−4
It follows that
2
−1 b
−2 c
0 0
0
(for F )
1+ c = 0
(for L )
−4 + a + b − 2c = 0
(for T )
2−b = 0
a = 0, b = 2, c = −1, and therefore
Π3 =
ρV 2
E
Check dimensions using MLT system:
ML−3 )( LT −1 )
(
ρV
=
= M L T
2
2
−1 −2
E
0 0
∴ ok
0
ML T
For Π 4:
Π4 = μd aV b E c
( FL T ) ( L )a ( LT ) ( FL ) = F L T
−1 b
−2
−2 c
0 0
0
(for F )
1+ c = 0
(for L )
−2 + a + b − 2c = 0
(for T )
1− b = 0
It follows that a = −1, b = 1, c = −1, and therefore
μV
Π4 =
Ed
Check dimensions using MLT system:
ML−1T −1 )( LT −1 )
(
=
= M L T
Ed
( L ) ( ML−1T −2 )
μV
0 0
Thus,
  ρV 2 μV 
=φ ,
,
 d E Ed 
d


δ
0
∴ ok
Problem 7.14
Because of surface tension, it is possible, with care, to support an object heavier than water
on the water surface as shown in the figure below. The maximum thickness, h , of a square
of material that can be supported is assumed to be a function of the length of the side of the
square, ; the density of the material, ρ ; the acceleration of gravity, g ; and the surface
tension of the liquid, σ . Develop a suitable set of dimensionless parameters for this
problem.
ℓ
h
Solution 7.14
h = L
h = f ( , ρ , g , σ )
ρ = FL−4T 2
 = L
g = LT −2
σ = FL−1
From the pi theorem, 5 − 3 = 2 pi terms required. Use , g , and ρ as repeating variables.
Thus,
Π1 = ha g b ρ c
and
( L )( L )a ( LT −2 ) ( FL−4T 2 ) = F 0 L0T 0
b
c
so that
c=0
1 + a + b − 4c = 0
−2b + 2c = 0
(for F )
(for L )
(for T )
It follows that a = −1, b = 0, c = 0, and therefore
Π1 =
h

which is obviously dimensionless.
For Π 2:
Π2 = σ a g b ρ c
( FL ) ( L )a ( LT ) ( FL T ) = F L T
−2 b
−1
1+ c = 0
−1 + a + b − 4c = 0
−2b + 2c = 0
−4
(for F )
(for L )
(for T )
It follows that a = −2, b = −1, c = −1, and therefore
Π2 =
σ
 gρ
2
2 c
0 0
0
Check dimensions using MLT system:
MT −2 )
(
= M L T
=
 2 g ρ ( L2 )( LT −2 )( ML−3 )
σ
0 0
Thus,
 σ 
h
=φ 2 

  gρ 
0
∴ ok
Problem 7.15
Under certain conditions, wind blowing past a rectangular speed limit sign can cause the
sign to oscillate with a frequency ω (see the figure below). Assume that ω is a function of
the sign width, b ; sign height, h ; wind velocity, V ; air density, ρ ; and an elastic constant, k ,
for the supporting pole. The constant, k , has dimensions of FL. Develop a suitable set of pi
terms for this problem.
SPEED
LIMIT
40
ω
Solution 7.15
ω = T
−1
ω = f ( b, h,V , ρ , k )
b = L
ρ = FL−4T 2
V = LT −1
h = L
k = FL
From the pi theorem, 6 − 3 = 3 pi terms required. Use b , V , and ρ as repeating variables.
Thus,
Π1 = ω baV b ρ c
and
(T ) ( L )a ( LT ) ( FL T ) = F L T
−1
−1 b
2 c
−4
so that
c=0
a + b − 4c = 0
−1 − b + 2c = 0
(for F )
(for L )
(for T )
It follows that a = 1, b = −1, c = 0, and therefore
Π1 =
ωb
V
Check dimensions:
T −1 ) ( L )
(
= L T
=
V
( LT −1 )
ωB
0
0
∴ ok
0 0
0
For Π 2:
Π2 = hbaV b ρ c
( L )( L )a ( LT −1 ) ( FL−4T 2 ) = F 0 L0T 0
b
c=0
1 + a + b − 4c = 0
−b + 2c = 0
c
(for F )
(for L )
(for T )
It follows that a = −1, b = 0, c = 0, and therefore
Π2 =
h
b
which is obviously dimensionless.
For Π 3:
Π3 = kbaV b ρ c
( FL )( L )a ( LT −1 ) ( FL−4T 2 ) = F 0 L0T 0
b
c
1+ c = 0
1 + a + b − 4c = 0
−b + 2c = 0
(for F )
(for L )
(for T )
a = −3, b = −2, c = −1, and therefore
It follows that
Π3 =
k
b V 2ρ
3
Check dimensions using MLT system:
k
bV ρ
3
2
=
ML2T −2
( L )( LT ) (ML )
3
−1 2
Thus,
h
k 
=φ , 3 2 
V
b bV ρ
ωb
−3
= M 0 L0T 0
∴ ok
Problem 7.16
A cone and plate viscometer consists of a cone with a very small angle α that rotates above
a flat surface as shown in the figure below. The torque, T, required to rotate the cone at an
angular velocity ω is a function of the radius, R ; the cone angle, α ; and the fluid viscosity,
μ ; in addition to ω . With the aid of dimensional analysis, determine how the torque will
change if both the viscosity and angular velocity are doubled.
R
ω
𝒯
α
Fluid
Solution 7.16
T== f ( R, α , μ , ω )
T == FL
μ = FL−2T
α = F 0 L0T 0
R = L
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containing T = ):
Π1 =
T=
μωR
3
=
FL
( FL T )(T ) ( L )
= F 0 L0T 0
= M 0 L0T 0
∴ ok
−2
−1
3
Check using MLT :
T=
μωR
=
3
(
ML2T −2
−1 −1
ML T
)(
T
−1
)(L)
3
The angle, α , can be used as Π 2 since it is dimensionless.
Thus,
T=
μω R3
= φ (α )
or
T== μω R 3φ (α )
It follows that if both μ and ω are doubled T= will increase by a factor of 4 .
ω = T −1
Problem 7.17
The pressure drop, Δp, along a straight pipe of diameter D has been experimentally studied,
and it is observed that for laminar flow of a given fluid and pipe, the pressure drop varies
directly with the distance, , between pressure taps. Assume that Δp is a function of D and ,
the velocity, V , and the fluid viscosity, μ . Use dimensional analysis to deduce how the
pressure drop varies with pipe diameter.
Solution 7.17
Δp = FL−2
Δp = f ( D, ,V , μ )
D = L
V = LT −1
 = L
μ = FL−2T
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containing Δp):
(
)
)(
FL−2 ( L )
ΔpD =
Π1 =
=
= F 0 L0T 0 ∴ ok
−2
−1
μV
FL T LT
(
)
Check using MLT :
(
)
ML−1T −2 ( L )
ΔpD =
= M 0 L0T 0
=
−1 −1
−1
μV
ML T
LT
(
)(
)
∴ ok
For Π 2(containing ):
Π2 =

D
which is obviously dimensionless. Thus,
ΔpD =

=φ 
μV
D
(1)
From the statement of the problem, Δp ∝  so that Eq. (1) must be of the form

ΔpD =
=K
D
μV
where K is some constant. It thus follows that
Δp ∝
for a given velocity.
1
D2
Problem 7.18
A cylinder with a diameter D floats upright in a liquid as shown in the figure below. When
the cylinder is displaced slightly along its vertical axis it will oscillate about its equilibrium
position with a frequency, ω . Assume that this frequency is a function of the diameter, D;
the mass of the cylinder, m; and the specific weight, γ , of the liquid. Determine, with the aid
of dimensional analysis, how the frequency is related to these variables. If the mass of the
cylinder were increased, would the frequency increase or decrease?
Cylinder
diameter = D
Liquid
Solution 7.18
ω = T
ω = f ( D, m, γ )
−1
γ  FL−3
m = FL−1T 2
D = L
From the pi theorem, 4 − 3 = 1 pi terms required.
By inspection:
Π1 =
ω
m
D
γ
T −1 ) FL−1T 2
(
=
= F L T
(L)
−3
0 0
0
FL
Check using MLT :
ω
m
D
γ
T −1 )
(
=
(L)
M
−2
ML T
−2
= M 0 L0T 0
∴ ok
Since there is only 1 pi term, it follows that
ω
m
D
γ
=C
where C is a constant. Thus,
ω = CD
γ
m
From this result it follows that if m is increased ω will decrease.
Problem 7.20
The weir shown in the figure below is used to measure the volume flowrate Q. The height H
is a measure of this flowrate. The weir has length L (perpendicular to the paper). Select and
include relevant fluid properties and find the appropriate dimensionless parameters.
H
Q
g
P
Density = ρ
Absolute viscosity = μ
Solution 7.20
Considering head above the weir H as the dependent variable and flowrate Q as an
independent variable.
H = f (Q , P , L , g , ρ , μ ) .
The dimensions of the parameters are
H L
3
Q
L
T
PL
LL
L
g 2
T
M
ρ 3
L
M
μ
LT
So n = 7, k = 3, and n − k = 4. Choose Q, P, and ρ as repeating variables. The first two
parameters are obvious
and
H
L
Π1 =
Π2 =
P
P
The third parameter is
Π3 = gQa P b ρ c
a
c
3
bM 
 L  L 
[Π3 ] =  2    ( L )  3 
 T  T 
L 
M:
c=0
L:
1 + 3a + b − 3c = 0
T:
−2 − a = 0
a = −2,
b = 5,
c = 0.
Π3 =
gP 5
Q2
The fourth parameter is
Π 4 = μQa P b ρ c
a
c
3
bM 
 M  L 
[ Π 4 ] =     ( L )  3 
 LT   T 
L 
M:
1+ c = 0
L:
−1 + 3a + b − 3c = 0
T:
−1 − a − 3c = 0
a = −1,
b = 1,
c = −1.
Π4 =
μP
ρQ
Problem 7.21
Experiments are conducted on a washing machine agitator. The relevant dimensional
parameters are the driving torque, T=; the oscillation frequency, f ; the angular velocity, ω ;
the number of paddles, N ; the paddle height, H ; and the paddle width, w . Specify the
relevant fluid properties and find the appropriate dimensionless parameters.
Solution 7.21
The relevant fluid properties are density ρ and viscosity μ .
Adding these, there are 8 variables T=, f , ω, N , H , w, ρ , and μ. Among them are 3
fundamental dimensions so we seek 5 dimensionless variables. N is already dimensionless
f
w
so we seek 4 more. The ratios and
are rather obvious. The remaining two could be a
ω
H
torque coefficient
Torque
2
Density × Velocity × Area × Moment arm
and a Reynolds Number Re =
=
T=
ρ (ω H ) H H
2
2
=
T=
ρω 2 H 5
Density × Velocity × Size ρ (ω H )H ρ (ω H )2
=
=
Viscosity
μ
μ
Problem 7.22
The input power, W , to a large industrial fan depends on the fan impeller diameter D, fluid
viscosity μ , fluid density ρ , volumetric flow Q, and blade rotational speed ω . What are the
appropriate dimensionless parameters?
Solution 7.22
Write the dimensional relation considering W as the dependent variable
W = f (Q, D, μ , ρ , ω ) .
Determine the maximum number of dimensionless parameters. There are six parameters
but η is already dimensionless so n = 5. Using the [M , L,T ] system
FL ML2
 3
W 
T
T
Q
L3
T
DL
FT
M
μ 2 
LT
L
M
ρ 3
L
1
ω
T
There are three fundamental dimensions ( M , L, T ) so we expect k = 3; D, μ , and ρ cannot
form a dimensions group so k = 3 and the maximum number of dimensionless parameters is
( 6 − 3 ) = 3.
Select D, ρ , and ω as repeating parameters. Combining D, ρ , and ω with W gives
 a ρ bω c.
Π1 = WD
The dimensional equation is
W  [ D ] [ ρ ] [ω ] = [M ] [ L ] [T ]
a
b
0
c
0
0
or
b
c
 ML2 
0
0
0
a M   1 
L
= [M ] [ L ] [T ]
[
]
 3 

3  T 
L   
 T 
Equating power of each dimension gives
M:
1+ b = 0
L:
2 + a − 3b = 0
T:
−3 − c = 0
Solving we obtain
b = −1,
c = −3,
a = −5.
The first dimensionless parameters is Π1 =
Combining D, ρ , and ω with Q gives
W
ρω 3D5
Π2 = QDa ρ bω c
The dimensional equation is
Q [ D ] [ ρ ] [ω ] = [M ] [ L ] [T ]
a
b
c
0
0
0
or
b
c
 L3 
0
0
0
a M   1 
L
= [M ] [ L ] [T ]
[
]
 




3
 L  T 
 T 
Equating power of each dimension gives
M:
b=0
L:
3 + a − 3b = 0
T:
−1 − c = 0
Solving we obtain
b = 0,
c = −1,
a = −3.
The second dimensionless group is Π 2 =
Q
ωD3
Combining D, ρ , and ω with μ gives
Π3 = μDa ρ bωc
The dimensional equation is
[ μ ][D]a [ ρ ]b [ω ]c = [M ]0 [ L ]0 [T ]0
or
b
c
0
0
0
 M  a M   1 
 LT  [ L ]  L3   T  = [M ] [ L ] [T ]
Equating power of each dimension gives
M:
L:
T:
Solving we obtain
b = −1,
1+ b = 0
−1 + a − 3b = 0
−1 − c = 0
c = −1,
The second dimensionless group is Π3 =
a = −2.
μ
ρωD2
Problem 7.23
Develop the appropriate dimensionless parameters for the period τ of transverse vibration
of a turbine rotor of mass m connected to a shaft of stiffness k = F / L and length . Other
relevant dimensional parameters are the eccentricity ε = L of the center of mass of the
rotor, the speed of rotation N = 1/ T of the shaft, and the amplitude A = L of the vibration.
The dimensions are Solution 7.23
τ = f ( m, k ,  , ε , N , A )
τ T
ε L
1
T
A L
mM
N
F
M
 2
L T
L
So n = 7, k = 3, and n − k = 4. Choose m, , N as repeating variables. Three parameters are
found by inspection.
ε
A
Π1 = τ N ,Π 2 = , and Π 3 = 2 .


k
The fourth parameter is
Π 4 = kma b N c
[ Π 4 ] =  2  ( M )a ( L )b (T )−c
M
M:
L:
T:
a = −1,
T 
1+ a = 0
b=0
−2 − c = 0
b = 0,
c = −2.
k
Π4 =
mN 2
Problem 7.24
A vapor bubble rises in a liquid. The relevant dimensional parameters are the liquid specific
weight, γ f ; the vapor specific weight, γ g ; bubble velocity, V ; bubble diameter, d ; surface
tension, σ ; and liquid viscosity, μ . Find appropriate dimensionless parameters.
Solution 7.24
(
V = f γ f ,γ g ,d ,σ , μ
)
dimensions are
L
d L
T
F
F
γf  3
σ
L
L
F
FT
γg  3
μ 2
L
L
So n = 6, k = 3 ( F , L, T ), and n − k = 3. An obvious dimensionless parameter is
V
Π1 =
γg
γf
Selecting V , d , and σ as repeating parameters, we combine them with γ f to get
γ f  [V ]a [ d ]b [σ ]c = F 0 L0T 0
 
or
a
c
 F  L 
bF 
0 0 0
 3   ( L )   = F L T
 L  T 
L
Equating powers of F , L, and T gives
F : 1+ c = 0
 a=0
L : −3 + a + b − c = 0 
 c = −1
 b = c − a + 3 = −1 − 0 + 3 = 2
T : −a = 0

Then
γ f d2
Π2 =
σ
Combining V , d , and σ with μ gives
[ μ ][V ]a [d ]b [σ ]c = F 0 L0T 0
or
a
c
bF 
 FT   L 
0 0 0
 2   (L)   = F L T
 L  T 
L
Equating power of F , L, and T give
F : 1+ c = 0
 a =1
L : −2 + a + b − c = 0  c = −1
 b = c − a + 2 = −1 − 1 + 2 = 0
T : 1− a = 0

Then
Π3 =
μV
σ
Problem 7.25
The speed of deep ocean waves depends on the wave length and gravitational acceleration.
What are the appropriate dimensionless parameters?
Solution 7.25
c = f ( λ , g ) where c =wave speed and λ = wave length. The dimensions are
c
L
L
, λ  L, and g  2 .
T
T
So n = 3 and k = 2, so n − k = 1. By inspection, the single dimensionless parameter is
Π=
c
gλ
Problem 7.26
A coach has been trying to evaluate the accuracy of a baseball pitcher. After 2 years of
studying, he proposes a function that can be presented as the accuracy of any pitcher:
Acc = f (V , a, m, ρ , p, z )
where Acc is the dimensionless accuracy, V is the velocity of the ball, a is the age of the
pitcher, m is the mass of the pitcher, ρ is the density of air where the game is played, p is the
pressure where the game is played (which varies with elevation), and z is the elevation
above sea level. Find the dimensionless groups for this function.
Solution 7.26
Acc = f (V , a, m, ρ , p, z )
Identity the relevant dimensional parameters
M
L
ρ 3
V
T
L
F
M
a T
p 2 
L
LT 2
zL
[ m] = M
Step 2. The number of physical parameters is n = 7 (including the dimensionless Acc). The
number of fundamental dimensions is k = 3 (for M , L, and T ) and since a, z , and p cannot
form a dimensionless groups. This gives n − k = 4 dimensionless groups. One obvious
dimensionless group is
Π1 = Acc
We now combine a, z , and p with V .
[V ][a ]a [ z ]b [ p]c = M 0 L0T 0
or
c
L
a
b M 
0 0 0
  (T ) ( L ) 
 =M L T .
2
T 
 LT 
Equating powers of M , L, and T give
c=0
b = c − 1 = −1
M : c=0

L : 1+ b − c = 0 

T : −1 + a − 2c = 0 

a = 2c + 1 = 1
Then
Π2 =
Va
z
We now combine a, z , and p with m.
[ m][a ]a [ z ]b [ p]c = M 0 L0T 0
or
c
M
( M )(T )a ( L )b  2  = M 0 L0T 0.
 LT 
Equating powers of M , L, and T gives
M : 1+ c = 0 
c = −1
L : b−c = 0 
b = c = −1

a = 2 c = −2
T : a − 2c = 0 

Then
Π3 =
m
a 2 zp
We now combine a, z , and p with ρ .
[ ρ ][a ]a [ z ]b [ p]c = M 0 L0T 0
or
c
a
b M 
M 
= M 0 L0T 0.
 3  (T ) ( L ) 
2
L 
 LT 
Equating powers of M , L, and T give
M : 1+ c = 0
c = −1


L : −3 + b − c = 0 
a = 2c = − 2
b = c+3 = 2
T : a − 2c = 0 

Then
Π4 =
ρ z2
a2 p
Problem 7.27
A viscous fluid is poured onto a horizontal plate as shown in the figure below. Assume
that the time, t; required for the fluid to flow a certain distance, d ; along the plate is a
function of the volume of fluid poured, V ; acceleration of gravity,g ; fluid density, ρ ; and
fluid viscosity, μ . Determine an appropriate set of pi terms to describe this process. Form
the pi terms by inspection.
Volume V
d
Solution 7.27
t = T
t = f ( d , V , g, ρ , μ )
g = LT −2
V = L3
d = L
ρ = FL−4T 2
From the pi theorem, 6 − 3 = 3 pi terms required.
By inspection, for Π1 (containing t):
Π1 =t
(
g (T )
=
d
( )
0 0
L3 )
(
=
= F L T
V
d
) = F L T ∴ ok
1
L 2
For Π 2 (containing V ):
Π2 =
1
−2 2
LT
0 0
0
0
∴ ok
)(
1
−2 2
LT
(L)
3
3
For Π 3(containing ρ and μ ):
Π3 =
3
gd 2
ρ
μ
=
(
−4
FL T
2
−2
)
3
(L)2
FL T
= F 0 L0T 0
Check using MLT system:
ρ
3
gd 2
μ
=
(
−3
ML
)(
1
−2 2
LT
) (L )2
(ML T )
−1 −1
Thus,
3 

g
V ρ gd 2

t
=φ 3 ,

d
d
μ


3
= M 0 L0T 0
∴ ok
μ = FL−2T
Problem 7.28
The velocity,c, at which pressure pulses travel through arteries (pulse–wave velocity) is a
function of the artery diameter, D; wall thickness, h ; the density of blood, ρ ; and the
modulus of elasticity, E , of the arterial wall. Determine a set of nondimensional parameters
that can be used to study experimentally the relationship between the pulse–wave velocity
and the variables listed. Form the nondimensional parameters by inspection.
Solution 7.28
c = LT
−1
c = f ( D, h, ρ , E )
ρ = FL−4T 2
h = L
D = L
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containing c):
Π1 =
c ρ
E
=
1
2 2
T
( LT )( FL ) = F L T
1
−2 2
( FL )
−1
−4
0 0
0
Check using MLT system:
c ρ
E
=
1
−3 2
ML
( LT )( ) = M L T
1
−1 −2 2
(ML T )
−1
For Π 2, let
Π2 =
h
D
which is obviously dimensionless. Thus,
c
ρ
h
=φ 
E
D
0 0
0
∴ ok
E = FL−2
Problem 7.29
As shown in the figure below, a jet of liquid directed against a block can tip over the block.
Assume that the velocity,V , needed to tip over the block is a function of the fluid density, ρ ,
the diameter of the jet, D, the weight of the block, W , the width of the block, b , and the
distance, d , between the jet and the bottom of the block. (a) Determine a set of
dimensionless parameters for this problem. Form the dimensionless parameters by
inspection. (b) Use the momentum equation to determine an equation for V in terms of the
other variables. (c) Compare the results of parts (a) and (b).
b
V
ρ
𝒲
D
d
Solution 7.29
(a)
V = LT −1
ρ = FL−4T 2
V = f ( ρ , D,W , b, d )
W = F
D = L
b = L
d = L
From the pi theorem, 6 − 3 = 3 pi terms required.
By inspection, for Π1 (containing V ):
Π1 = VD
 FL−4T 2 
 = F 0 L0T 0
= LT −1 ( L ) 


W
F


ρ
(
)
Check using MLT :
VD
 ML−3 
 = M 0 L0T 0
= LT −1 ( L ) 
 MLT −2 
W


ρ
(
)
∴ ok
For Π 2, let:
Π2 =
b
d
Π3 =
d
D
and for Π 3:
and both Π 2and Π 3 are obviously dimensionless.
b
F
Thus,
VD
ρ
b d 
=φ , 
W
d D
(b) For impending tipping around 0
 M0 = 0
𝒲
d
0
So that
b
Fd = W  
(1)
2
From momentum considerations using cv shown
+
ˆ dA =  Fx
⎯⎯→ ρ nVm
ρV 2 A = F
cv
Thus, from Eq. (1)
F
( ρV A) ( d ) = W  b2 
V
2
A ~ area
so that
V=
W (b)
2 ρ Ad
y
and with A =
V=
π
4
x
D
2
2Wb
πρ dD 2
(2)
(c) From part (a)
V=
W
ρD
b d 

d D
φ ,
2 
Eq. (2) can be written as
V=
W   2  b  
    
ρ D 2   π   d  
(3)
It follows by comparing Eqs. (2) and (3) that
b d 
 2  b 
 =   
d D
 π  d 
φ ,
d
b d 
So that φ  ,  is actually independent of .
D
d D
Problem 7.30
By inspection, arrange the following dimensional parameters into dimensionless
parameters: (a) kinematic viscosity, v, length, , and time, t; and (b) volume flowrate, Q,
pump diameter, D, and pump impeller rotation speed, N .
Solution 7.30
(a) The dimensional parameters and their dimensions are:
v
L2
T
L
Πa =
vt
2
t T
(b) The dimensional parameters and their dimensions are:
Q
L3
T
DL
N
1
T
Πb =
Q
ND3
Problem 7.31
A screw propeller has the following relevant dimensional parameters: axial thrust, F ;
propeller diameter, D; fluid kinematic viscosity, v; fluid density, ρ ; gravitational
acceleration, g ; advance velocity, V ; and rotational speed, N . Find appropriate
dimensionless parameters to present the test data.
Solution 7.31
F = f (V , N , D, v, ρ , g )
F
L2
T
M
ρ 2
L
L
g 2
T
ML
T
L
V
T
1
N
T
ν
2
DL
n = 7, k = 3 , and n − k = 4 . Assuming that the propeller operates near a free surface in
water, the relevant dimensionless parameters must include
Reynolds number
Froude number
Velocity ratio
We will use the propeller tip velocity ( ND ) as a reference velocity and diameter D as a
reference length. Then
Π1 = Re =
( ND ) D = ND2
v
v
, Π2 = F =
ND N D
V V
=
, Π3 = =
.
U VD
gD
g
The remaining parameter to form into a dimensionless group is the thrust force F . It can
be made into a force coefficient similar to a lift or drag coefficient. Continuing to use ND
for velocity and D for length gives
Π4 =
F
ρ ( ND ) D2
2
=
F
ρ N 2D4
.
Problem 7.32
Shown in the following table are several flow situations and the associated characteristic
velocity, size, and fluid kinematic viscosity. Determine the Reynolds number for each of the
flows and indicate for which ones the inertial effects are small relative to viscous effects.
Flow
Velocity (ft/s)
Size (ft)
Airplane
Mosquito
Syrup on pancake
Blood in capillary
400
0.02
0.03
0.001
60
0.015
0.12
5 × 10−5
Kinematic viscosity
(ft2/s)
4.3 × 10−4
1.6 × 10−4
2.5 × 10−3
1.2 × 10−5
Solution 7.32
Re =
V
so that for the airplane
v
ft 

 400  ( 60 ft )
s
Re = 
= 5.58 × 107  1
2
ft
4.3 × 10 −4
s
Since Re  inertia force/viscous force, it follows that inertial effects are not small relative to
viscous effects for the airplane.
The table below shows the results for all 4 flows.
Airplane
Mosquito
Syrup
Blood
Re
5.58 × 107
1.88
1.44
4.17 ×10 −3
Inertial effects small relative to viscous effects?
No
Both important
Both important
Yes
Problem 7.33
A large, hot plate hangs vertically in a room. Heat transfer from the plate to the air near it
causes the air density to decrease. This lighter air rises upward with a velocity proportional
to the group
(
)
1/2
αT gL Tp − T f  ,


where αT is the coefficient of thermal expansion [units =( °R )−1 or K −1], g is the acceleration
of gravity, L is the plate length, and T p and Tf are the temperatures of the plate and fluid,
respectively, in °R or K . The resulting heat transfer process depends on the Grashof
Number, defined by
Gr =
ρ 2αT g (Tp − T f ) L3
μ2
,
where ρ is density and μ is dynamic viscosity. Show that G is dimensionless. How is G
related to the common dimensionless parameters of fluid mechanics? Is it equivalent to a
combination, power, or product of one or more of the common dimensionless parameters
Solution 7.33
2
( )
 M   1  L 
3
 3    2  (θ ) L
θ
 Gr  1 .
Gr   L    T 2
M 


 LT 
Since
(
)
1
V ∝ αT gL Tp − T f  2 ,


then
2
ρ 2V 2 L2  ρVL 
2
Gr ∝
=
 = Re
2
μ
 μ 
where Re is the Reynolds number. Then
Gr ∝ Re2 , a power.
Problem 7.34
Develop the Weber number by starting with estimates for the inertia and surface tension
forces.
Solution 7.34
Inertia force  ρV 2  2
and
Surface tension force  σ 
then
Inertia force
ρV 2  2 ρV 2 

=
σ
σ
Surface tension force
This is Weber number, W =
ρV 2 
σ
Problem 7.35
Develop the Froude number by starting with estimates of the fluid kinetic energy and fluid
potential energy.
Solution 7.35
Kinetic Energy  mV 2 = ρ  3V 2 ( = typical length)
and
Potential Energy  mg  = ρ  3 g   ρ g  4,
then
Kinetic Energy
ρ 3V 2 V 2

=
= F2.
4
Potential Energy
g
ρ g
Froude number is the square root, Fr =
V2
V
=
g
g
Problem 7.36
The following dimensionless groups are often used to present data on centrifugal pumps:
W
Q
gH
ξ
=
flow coefficient φ =
,
head
coefficient
,
power
coefficient
, efficiency
ψ
=
ω 3D5
ω D3
ω 2D2
ω Q
ω Q
ρ gQH
D
=
, specific speed Ns =
,
specific
diameter
. Show that the last
η=
s
W
( gH )3/4
( gh )1/4
three groups can be formed from combinations of the first three groups.
Solution 7.36
1/2
 Q 


1/2
φ
(Q1/2 ) × (ω )6/4 D 6/4
ω (Q )1/2
 ω D3 
=
=
=
= Ns
ψ 3/4  gH 3/4 (ω )1/2 ( D )3/2 ( gH )3/4 ( gH )3/4
 2 2
ω D 
Q
gH
× 2 2
3
QgH × ρω 3D5 ρQgH
φψ ω D ω D
=
=
=
=η
W
W
ξ
ω 3D5 ×W
ρω 3D5
1/4
 gH 
 2 2
1/4
1/2
3/2
1/4
1/4
ψ
 ω D  = (ω ) ( D ) ( gH ) = D( gH ) = D
=
s
1/2
φ 1/2
(Q1/2 ) × (ω )1/2 D1/2
(Q )1/2
 Q 


 ω D3 
Problem 7.37
The dimensional parameters used to describe the operation of a ship or airplane propeller
(sometimes called a screw propeller) are rotational speed, ω ; diameter, D; fluid density, ρ ;
speed of the propeller relative to the fluid, V ; and thrust developed, T . The common
dimensionless groups are called the thrust coefficient and the advance ratio. Propose
appropriate definitions for these groups.
Solution 7.37
A number of different definitions could be proposed. The most common are:
Thrust Coefficient =
Thrust Force
Thrust Force
=
Momentum Flow Density × Velocity 2 × Area
CT =
T
ρ × (ω D ) × D
2
Advance Ratio =
2
=
T
ρω 2 D 4
Forward Speed
Propellor (Blade) Speed
AR =
V
ωD
Problem 7.38
The pressure rise, Δp = p2 − p1, across the abrupt expansion of the figure below through
which a liquid is flowing can be expressed as
Δp = f ( A1, A2 , ρ ,V1 )
where A1 and A2 are the upstream and downstream cross-sectional areas, respectively,ρ is
the fluid density, and V1 is the upstream velocity. Some experimental data obtained with
A2 = 1.25ft 2, V1 = 5.00 ft/s, and using water with ρ = 1.94 slugs/ft 3 are given in the following
table:
A1 (ft 2 )
2
Δp (lb/ft )
0.10
3.25
0.25
7.85
0.37
10.3
0.52
11.6
0.61
12.3
Plot the results of these tests using suitable dimensionless parameters. With the aid of a
standard curve fitting program determine a general equation for Δp and use this equation to
A
predict Δp for water flowing through an abrupt expansion with an area ratio 1 = 0.35 at a
A2
velocity V1 = 3.75ft/s.
p2
p1
V1
Area = A1
Area = A2
Solution 7.38
Δp = FL−2
ρ = FL−4T 2
A1 = L2
A2 = L2
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containingΔp):
Π1 =
Δp=
=
2
ρV1
FL−2
( FL−4T 2 )( LT )
−1 2
= F 0 L0T 0
Check using MLT :
Δp=
=
2
ρV1
For Π 2 (containing A1 andA2 ):
ML−1T −2
(ML )( LT )
−3
Π2 =
−1
2
= M 0 L0T 0
A1
A2
which is obviously dimensionless. Thus,
 A1 
φ
=
 
ρV12
 A2 
Δp=
∴ ok
V1 = LT −1
Using the data given, it follows that:
0.067
0.162
Δp / ρV12
A1 / A2
0.080
0.200
0.212
0.239
0.254
0.296
0.416
0.488
A plot of these data is shown below:
0.3
0.25
Pi 1
0.2
0.15
0.1
0.05
0
0
0.1
0.2
0.3
0.4
0.5
Pi 2
The curve drawn on the graph is a 2-nd order polynomial giving the equation
2
 A1 
A 
Δp=
= −1.10   + 1.07  1  − 0.0103
2
ρV1
 A2 
 A2 
A1
ft
= 0.35 and V1 = 3.75 with water (ρ = 1.94slugs/ft 3 )
A2
s
2
slugs  
ft 
lb

2
Δp =  1.94 3   3.75   −1.10 ( 0.35 ) + 1.07 ( 0.35 ) − 0.0103 = 6.26 2

s 
ft  
ft

Thus, for
Problem 7.39
The pressure drop, Δp, over a certain length of horizontal pipe is assumed to be a function
of the velocity, V , of the fluid in the pipe, the pipe diameter, D, and the fluid density and
viscosity, ρ and μ . (a) Show that this flow can be described in dimensionless form as a
“pressure coefficient, ” Cp = Δp / (1/ 2 ρV 2 ) that depends on the Reynolds number,
Re = ρVD / μ. (b) The following data were obtained in an experiment involving a fluid with
ρ = 2 slugs/ft3, μ = 2 × 10−3 lb⋅ s/ft2, and D = 0.1 ft. Plot a dimensionless graph and use a
power law equation to determine the functional relationship between the pressure
coefficient and the Reynolds number. (c) What are the limitations on the applicability of
your equation obtained in part (b)?
Δp (lb/ft2)
192
704
1088
1280
V (ft/s)
3
11
17
20
Solution 7.39
Δp = f (V , D, ρ , μ )
(a)
Δp = FL−2
V = LT −1
ρ = FL−4T 2
D = L
μ = FL−2T
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1:
Π1 =
Δp=
ρV
=
2
FL−2
= F 0 L0T 0
∴ ok
= M 0 L0T 0
∴ ok
) ( L ) = F 0 L0T 0
∴ ok
( FL T )( LT )
−4
−1 2
2
Check using MLT system:
Δp=
ρV
=
2
ML−1T −2
(ML )( LT )
−1 2
−3
For Π 2:
(
−4
2
)(
FL T
LT
ρVD
Π2 =
=
μ
FL−2T
(
−1
)
Check using MLT system:
(
)(
)
)
LT ( L )
ρVD ML
=
= M 0 L0T 0
−
−
1
1
μ
ML T
−3
(
−1
∴ ok
Thus,
 ρVD 
= φ 

ρV
 μ 
Δp=
2
Since φ is an unknown function, a factor of
can be included in Π1 (if desired) so that
 ρVD 
=
φ


1/ 2 ρV 2
 μ 
Δp
Thus,
C p = φ ( Re )
where Cp is the pressure coefficient and Re the Reynolds number.
(b) Using the data given,
Δp
Δp
Δp
=
=
Cp =
2
slugs
1/ 2 ρV 2
( 0.5 )  2 3 V 2 V
 ft 
and
 slugs 
2  3  (V )( 0.1ft )
ρVD
Re =
=  ft 
= 100V
lb ⋅ s
μ
2 ×10 −3 2
ft
Cp
Pressure coefficient,
Tabulated values for Cp and Re and a plot of the data are shown below.
V (ft/s)
p (psf)
Re
Cp
3
11
17
20
192
704
1090
1280
300
1100
1700
2000
21.3
5.82
3.77
3.20
25.0
20.0
15.0
10.0
5.0
0.0
Cp = 6387 Re–1.00
0
500
1000 1500 2000 2500
Reynolds number, Re
The power law relationship is
6387
Cp =
Re
(1)
(c) Based on the variable used and the given data, the empirical relationship, Eq. (1), would
only be applicable in the Reynolds number range
300 ≤ Re ≤ 2000
Note: Although the equation might be valid outside this range, results should not be
extrapolated beyond the range of data used.
Problem 7.40
The pressure drop across a short hollowed plug placed in a circular tube through which a
liquid is flowing (see the figure below) can be expressed as
Δp = f ( ρ ,V , D, d )
where ρ is the fluid density, and V is the mean velocity in the tube. Some experimental data
obtained with D = 0.2 ft, ρ = 2 slugs/ft 3, and V = 2 ft/s are given in the following table:
d (ft)
2
Δp (lb/ft )
0.06
493.8
0.08
156.2
0.10
64.0
0.15
12.6
Plot the results of these tests, using suitable dimensionless parameters, on a log–log scale.
Use a standard curve-fitting technique to determine a general equation for Δp. What are the
limits of applicability of the equation?
Δp
V
D
d
Solution 7.40
Δp = FL−2
ρ = FL−4T 2
V = LT −1
d = L
D = L
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containing Δp):
Π1 =
Δp=
ρV
=
2
FL−2
( FL T )( LT )
−4
−1 2
2
= F 0 L0T 0
Check using MLT :
Δp=
ρV
=
2
ML−1T −2
(ML )( LT )
−1 2
−3
= M 0 L0T 0
For Π 2 (containing D and d ):
Π2 =
D
d
(which is obviously dimensionless). Thus,
Δp=
D
=φ 
d 
ρV
2
For the data given:
3.33
D/d
2
61.7
Δp / ρV
2.50
19.5
2.00
8.00
1.33
1.58
∴ ok
A log–log plot of these data is shown below.
Pi 1
100
10
1
1
10
Pi 2
Since the data plot as a straight line on a log–log plot, the equation for the data is of the
form
Π1 = aΠb2
D
Δp
where Π1 =
and
. A power law fit of the data gives
Π
=
2
d
ρV 2
a = 0.505 and b = 3.99
Thus,
Δp
3.99
D
= 0.505  
2
d 
ρV
This equation is applicable over the range of data 1.33 ≤
D
≤ 3.33.
d
Problem 7.41
As shown in the figure below, a rectangular barge floats in a stable configuration provided
the distance between the center of gravity, CG, of the object (boat and load) and the center
of buoyancy, C, is less than a certain amount, H . If this distance is greater than H , the boat
will tip over. Assume H is a function of the boat’s width, b , length, , and draft, h . (a) Put
this relationship into dimensionless form. (b) The results of a set of experiments with a
model barge with a width of 1.0 m are shown in the table. Plot these data in dimensionless
form and determine a power–law equation relating the dimensionless parameters.
ℓ (m)
h (m)
H (m)
2.0
4.0
2.0
4.0
2.0
4.0
0.10
0.10
0.20
0.20
0.35
0.35
0.833
0.833
0.417
0.417
0.238
0.238
CG
H
h
C
b
Solution 7.41
(a)
H = f ( b, , h )
From the pi theorem, 4 − 1 = 3 pi terms required.
By inspection:
H
h 
=φ , 
b
b b
h/b
H/b
ℓ/b
0.10
0.10
0.20
0.20
0.35
0.35
0.833
0.833
0.417
0.417
0.238
0.238
2.0
4.0
2.0
4.0
2.0
4.0
H/b
All of the pi terms are obviously dimensionless.
H h

(b) For the data given, tabulated values for , , and are shown below.
b
b b
1
0.8
0.6
0.4
0.2
0
0.00
H/b = 0.0833 (h/b)–1.00
0.10
0.20
h/b
0.30
0.40
An inspection of these data reveals that
H
does not depend on , i.e., the same value of
b
b
H
is obtained for different values of . Thus,
b
b
H
h
=φ 
b
b
and from the plot of the data, using a power–law equation
H
h
= 0.0833  
b
b
−1.00
Problem 7.42
The time, t, it takes to pour a certain volume of liquid from a cylindrical container depends
on several factors, including the viscosity of the liquid. Assume that for very viscous liquids
the time it takes to pour out two-thirds of the initial volume depends on the initial liquid
depth, , the cylinder diameter, D, the liquid viscosity, μ , and the liquid specific weight, γ .
The data shown in the following table were obtained in the laboratory. For these tests,
 = 45 mm, D = 67 mm, and γ = 9.60 kN/m 3 . (a) Perform a dimensional analysis, and based
on the data given, determine if variables used for this problem appear to be correct. Explain
how you arrived at your answer. (b) If possible, determine an equation relating the pouring
time and viscosity for the cylinder and liquids used in these tests. If it is not possible,
indicate what additional information is needed.
11
17
39
61
107
μ (N · s/m2)
t(s)
15
23
53
83
145
μ = FL−2T
γ = FL−3
Solution 7.42
t = f ( , D, μ , γ )
t = T
 = L
D = L
From the pi theorem, 5 − 3 = 2 pi terms required.
By inspection, for Π1 (containing t):
Π1 =
Check using MLT system:
tγ D
μ
=
tγ D
μ
=
(T ) ( FL−3 ) ( L )
( FL T )
−2
(T ) ( ML−2T −2 ) ( L )
(
−1 −1
ML T
)
= F 0 L0T 0
= M 0 L0T 0
∴ ok
For Π 2 (containing ):
Π2 =

D
which is obviously dimensionless. Thus,
tγ D

=φ 
(1)
μ
D
 45 mm
= 0.672 (a constant).
For the data given =
D 67 mm
tγ D

Thus, from Eq. (1) with a constant it follows that
= constant. For the data given:
μ
D
tγ D
877
870
874
875
872
μ
Since Π1 is essentially constant over the range of the experimental data the variables used
for the problem appear to be correct.
(b) The average value for Π1 is 874 so that
tγ D
μ
= 874
and therefore
t=
874
874 μ
μ=
γD

3 N 
67 ×10 −3 m
 9.6 ×10
3
m 

t = 1.36 μ
with t in seconds when μ is in units
(
N ⋅s
)
.
m2

kN
Note that this restricted equation is only valid for = 0.672, D = 67 mm, and  = 9.60 3
D
m
with two-thirds of the initial volume being poured.
Problem 7.43
In order to maintain uniform flight, smaller birds must beat their wings faster than larger
birds. It is suggested that the relationship between the wingbeat frequency, ω , beats per
second, and the bird’s wingspan, , is given by a power law relationship, ω   n . (a) Use
dimensional analysis with the assumption that the wingbeat frequency is a function of the
wingspan, the specific weight of the bird, γ , the acceleration of gravity, g , and the density of
the air, ρa , to determine the value of the exponent n. (b) Some typical data for various birds
are given in the following table. Do these data support your result obtained in part (a)?
Provide appropriate analysis to show how you arrived at your conclusion.
Bird
Wingspan (m)
Wingbeat frequency
(beats/s)
0.28
0.36
0.46
1.00
1.50
1.80
5.3
4.3
3.2
2.2
2.6
2.0
Purple martin
Robin
Mourning dove
Crow
Canada goose
Great blue heron
Solution 7.43
(a) Given ω = f ( , γ , g , ρa ) so that k − r = 5 − 3 = 2 or Π1 = φ ( Π 2 )
γ = ML−2T −2
g = LT −2
ω = T −1
 = L
Thus, consider
(
Π1 = ωρaa g b c = T −1 ML−3
So that
ρa = ML−3
) ( LT ) Lc = M a L a b cT
a
−2 b
−3 + +
−1−2 b
= M 0 L0T 0
M : a=0
L:
−3a + b + c = 0
T:
−1 − 2 b = 0
1
1
which gives a = 0, b = − , c =
2
2
Thus:
Π1 = ω

g
For Π 2:
(
)(
Π 2 = γρaa g b c = ML−2T −2 ML−3
So that
M : 1+ a = 0
L : −2 − 3a + b + c = 0
T:
−2 − 2 b = 0
These equations give a = −1, b = −1, and c = 0
) ( LT ) L = M L
a
−2 b
c
1+ a −2 −3a + b +c
T −2−2b
Thus,
Π2 =
Therefore, ω
γ
g ρa
 γ 
 γ 

g
=φ
=φ
 or ω

g

 g ρa 
 g ρa 
which indicates that ω  
−
1
2 . That is ω   n where n = − 1
2
(b) The given data are plotted below and a power law curve fit is applied, with the results
beats
ω = 2.62 −0.455, where ω 
when   m.
s
The obtained power, −0.455, is very close to that predicted by dimensional methods, −0.500.
6
5
ω, beats/s
4
3
2
y = 2.6207x–0.4553
1
0
0
0.5
1
ℓ = wingspan, m
1.5
2
Problem 7.44
A 250-m- long ship has a wetted area of 8000 m2 . A
1
-scale model is tested in a towing
100
tank with the prototype fluid, and the results are:
Model velocity (m/s)
0.57
1.02
1.40
Model drag (N)
0.50
1.02
1.65.
Calculate the prototype drag at 7.5 m /s and 12.0 m /s.
Solution 7.44
The dimensionless relationship for drag of surface vessels is
CD = f ( Fr,Re ).
As explained in the text, it is generally impossible to match both Reynolds number and
Froude number in a scale model test. For surface vessels, the Froude number is the primary
similarity parameter which must be satisfied and the Reynolds number is a secondary
similarity parameter. Therefore, we use
CD = f ( F )
Now for the model test,
CDm =
m
1
ρ mVm2 Am
2
,
2
(
)
 1 
2
2
where Am is the wetted area, and Am = 
 8000 m = 0.80 m ,
 100 
Vm
Vm
Vm
.
Fr =
=
=
m
g m
m  1 

4.95
 ( 250 m )
 9.81 2  
s
100
s 


Also
CDm =
The test data give
Vm (m/s)
m (N)
Fr
CD
m
(
)
1
kg 
2
2
1000 3  0.8 m Vm
2
m 
0.57
0.50
0.115
0.00835
=
m
400Vm2
1.02
1.02
0.206
0.00245
.
1.40
1.65
0.283
0.00210
Plotting this data gives
.004
.003
.002
0
0.100
0.200
0.300
The required prototype velocities give
Fr =
m

 7.5 
s 

=
= 0.151
g p
m

 9.81 2  ( 250 m )
s 

Vp
and
m

12.0 

Vp
s 

Fr =
=
= 0.242
g p
m

 9.81 2  ( 250 m )
s 

The above figure gives C D = 0.00307 and CD = 0.00225. Then
(
2
)
kg 
m
1
 1 
2
2 
1 = CD  ρ AV  = 0.00307   1000 3  8000 m  7.5
,
s 
m 
2

 2 

m
= 691000 N, or
1 = 691 kN at 7.5
s
and
(
)
kg 
m
1
 1 
2
2 
2 = CD  ρ AV  = 0.00225   1000 3  8000 m  12.0

s 
2

 2 

m 
= 1300000 N, or
2 = 1300 kN at 12.0
m
s
2
Problem 7.45
A student is interested in the aerodynamic drag on spheres. She conducts a series of wind
tunnel tests on a 10-cm- diameter sphere. The air in the wind tunnel is at 50 C and 101.3kPa.
She presents the results of her tests in the form of a correlation,
= 0.0015V 1.92,
where is the drag force in Newton and V is the velocity in m/s. Another student realizes
that the results would have been more effectively presented in terms of dimensionless
parameters, say,
Π1 = f ( Π 2 ) = C Π a2,
where Π1 and Π 2 are dimensionless parameters and C and a are constants. What are the
most appropriate dimensionless parameters (Π1 and Π 2) and the corresponding values of C
and a ? What would be the drag on a 2-cm-diameter sphere placed in a 1-m/s, 20 C water
stream?
Solution 7.45
Assuming Weber and Froude number effects are not relevant,
CD = f ( Re, α )
Since a sphere has no angle of attack,
CD = f ( Re )
or
D
1
ρ D 2V 2
2
for a sphere of diameter D . The data suggest
 ρVD 
f

 μ 
CD = C Rea
so Π1 = C D and Π 2 = Re .
This gives
 ρVD 
=C 

2 2
ρD V
 μ 
a
or
=
ρ D2V 2
2
a
 ρVD  C ρ1+a ( DV )2+a
C
 =
2μ a
 μ 
(2)
Comparing Eqs. (1) and (2) gives 2 + a = 1.92 or a = −0.08 and
C ρ1+a D2 +a
2μ a
C=
= 0.0015
0.0030 μ a
ρ
1+ a
D
2+a
=
giving
0.0030
ρ
0.92
D1.92 μ 0.08
Using data for 50 °C air,
C















C



 





(






)
−
C
Then CD = 0.549 Re−0.08 .
For the D = 0.02 m sphere in 20 C water with V = 1
m
s
 m
1
(0.02 m)
VD  s 
Re =
=
= 2.0 ×10 4 ,
2
ν

−6 m 
1.0 ×10

s 

CD = 0.549(2.0 ×10 4 )−0.08 = 0.249,
2
kg 
2 m

998 3  ( 0.249 )( 0.02 m ) 1 
2 2

ρCD D V
m 
 s  ,
=
=
2
 kg ⋅ m 
2

 N ⋅ s2 
= 0.0497 N



Problem 7.46
Air at 80°F is to flow through a 2-ft pipe at an average velocity of 6 ft/s. What size pipe
should be used to move water at 60 °F and average velocity of 3 ft/s if Reynolds number
similarity is enforced?
Solution 7.46
For Reynolds number similarity,
Reair = Rewater ,
or
 VD 
 VD 

 =

 ν air  ν  water
Thus,
ν
 V

Dwater =  water  air  Dair,
 ν air  Vwater 
where
ν water = 1.210 ×10
−5 ft
60  F
ν air
= 1.69 ×10−4
80  F
2
s
and
ft 2
s
Hence,
2

−5 ft   ft 
 1.210×10
 6
s   s  ( 2 ft ) = 0.286 ft
Dwater = 
2  ft 

−4 ft   3

 1.69×10

s  s 

Problem 7.47
You are to conduct wind tunnel testing of a new football design that has a smaller lace
height than previous designs. It is known that you will need to maintain Re and St
similarity for the testing. Based on standard college quarterbacks, the prototype parameters
are set at V = 40 mph and ω = 300rpm, where V and ω are the speed and angular velocity of
the football. The prototype football has a 7-in. diameter. Due to instrumentation required
to measure pressure and shear stress on the surface of the football, the model will require a
length scale of 2 :1 (the model will be larger than the prototype). Determine the required
model free stream velocity and model angular velocity.
Solution 7.47
Let ( )m and ( )p denote the model and prototype, respectively.
For Reynolds number similarity, Re m = Rep, or
Vm Dm
νm
Vm =
=
Vp Dp
νp
, so that with ν m ν p (i.e., same air properties)
1
Vp =   ( 40 mph ) = 20 mph , since Dm = 2Dp.
Dm
2
Dp
Thus,
Vm = 20
mi  5280 ft   1hr 
ft
 =29.3


hr  mi   3600 s 
s
For Strouhal number similarity, St m = St p, or
ωmDm
Vm
=
ω p Dp
Vp
, where ω p = 300 rpm
Hence,
ωm =
 20 mph   1 
Vm Dp
ωp = 
   ( 300 rpm ) = 75 rpm
V p Dm
 40 mph   2 
Problem 7.48
A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard
sea-level air, while the prototype will be operated in seawater. Determine the speed of the
prototype to ensure Reynolds number similarity.
Solution 7.48
Let ( ) m and ( ) p denote the model and prototype, respectively.
Thus, Re m = Re p, or
Vm  m
νm
=
Vp  p
νp
, where  m
1
p
15
Hence,
ν    p 
ν m 
Vm =  m  
V
=
15

V
 ν p    m  p
ν p  p




Also,
ν m = 1.57 ×10 −4
ft 2
s
and ν p = 1.26 × 10−5
ft 2
so that
s
2

−4 ft 
 1.57 × 10

s V = 187V
Vm = 15 
p
p
2

−5 ft 
 1.26 × 10

s 

Thus,
ft
180
Vm
s = 0.963 ft
Vp =
=
187
187
s
Problem 7.49
The drag characteristics of a torpedo are to be studied in a water tunnel using a 1: 5 scale
model. The tunnel operates with fresh water at 20 C , whereas the prototype torpedo is to
be used in seawater at 15.6 C . To correctly simulate the behavior of the prototype moving
with a velocity of 30 m/s, what velocity is required in the water tunnel?
Solution 7.49
For dynamic similarity, the Reynolds number must be the same for model and prototype.
Thus,
Vm Dm
νm
=
VD
Vm
=
ν
so that
Vm =
ν
D
V
Dm
Since, ν m (water @ 20 C ) = 1.004 ×10 −6
and
m2
m2
, and ν (seawater @ 15.6 C ) = 1.17 ×10 −6
s
s
D
= 5, it follows that
Dm
2

−6 m 
1.004×10

s 
m
m

Vm =
(5 )  30  = 129
2
s 
s


−6 m 
1.17×10

s 

Problem 7.50
For a certain fluid flow problem, it is known that both the Froude number and the Weber
number are important dimensionless parameters. If the problem is to be studied using a
1:15 scale model, determine the required surface tension scale if the density scale is equal
to 1. The model and prototype operate in the same gravitational field.
Solution 7.50
For dynamic similarity,
Vm
V
=
gm  m
g
(Froude number similarity)
and
ρ mVm2  m ρV 2 
=
σm
σ
(Weber number similarity)
To satisfy Froude number similarity (with g = gm),
Vm
m
=

V
and therefore for Weber number similarity
ρ V  
ρ  
ρ  
σ
= m m m = m m m = m m
ρ V  
ρ    
ρ   
σm
ρ

1
and m = 1,
Thus, with m =
ρ

15
2
σm
 1 
= (1) 
= 4.44 ×10−3

σ
 15 
2
Problem 7.51
The fluid dynamic characteristics of an airplane flying 240 mph at 10,000ft are to be
investigated with the aid of a 1: 20 scale model. If the model tests are to be performed in a
wind tunnel using standard air, what is the required air velocity in the wind tunnel? Is this a
realistic velocity?
Solution 7.51
For dynamic similarity, Reynolds number must be the same for model and prototype.
Thus,
ρ mVm  m ρV 
=
μm
μ
So that
Vm =
μm ρ 
V
μ ρm  m
(1)
Since,
μ = 3.534 × 10 −7
lb ⋅ s
μ m = 3.74 × 10 −7
lb ⋅ s
and
ft
2
ft
2
;
ρ = 1.756 × 10 −3
slugs
;
ρ m = 2.38 × 10 −3
slugs
ft 3
ft 3
(see the table below)

= 20 it follows from Eq. (1) That
m
slugs 

−7 lb ⋅ s  
1.756 × 10 −3 3 
 3.74 ×10
2  
ft  
ft  20 240 mph = 3750 mph
Vm = 
( )(
)

−7 lb ⋅ s  
−3 slugs 
 3.534 ×10
  2.38 ×10

ft 2  
ft3 

No, it is not a realistic velocity—much too high.
Problem 7.52
If an airplane travels at a speed of 1120km/hr at an altitude of 15km , what is the required
speed at an altitude of 8km to satisfy Mach number similarity? Assume the air properties
correspond to those for the U.S. standard atmosphere.
Solution 7.52
For Mach number similarity,
V 
V 
= 
 
 c 15km  c 8km
(1)
The speed of sound can be calculated from the equation
c = kRT
and for air, k = 1.40, R = 286.9
J
.
kg ⋅ K
For the Standard Atmosphere at 15km altitude,
T = −56.50 C + 273.15 = 216.7 K
and at 8km
T = −36.94 C + 273.15 = 236.2 K
Thus, at 15km altitude

c15km =
(1.40 )  286.9
c8km =
(1.40 )  286.9

J 
m
 ( 216.7 K ) = 295
kg ⋅ K 
s
and at 8km


J 
m
 ( 236.2 K ) = 308
kg ⋅ K 
s
From Eq. (1)
m

 308 s  
km 
km
1120
=1170
V8km =
V15km = 



hr 
hr
c15km
 295 m  
s 

c8km
Problem 7.53
Modeling parachutes in a water tunnel The first use of a parachute with a free-fall jump from
an aircraft occurred in 1914, although parachute jumps from hot-air balloons had occurred
since the late 1700s. In more modern times, parachutes are commonly used by the military and
for safety and sport. It is not surprising that there remains interest in the design and
characteristics of parachutes, and researchers at the Worcester Polytechnic Institute have been
studying various aspects of the aerodynamics associated with parachutes. An unusual part of
their study is that they are using small-scale parachutes tested in a water tunnel. The model
parachutes are reduced in size by a factor of 30–60 times. Various types of tests can be
performed, ranging from the study of the velocity fields in the wake of the canopy with a
steady free-stream velocity to the study of conditions during rapid deployment of the canopy.
According to the researchers, the advantage of using water as the working fluid, rather than
air, is that the velocities and deployment dynamics are slower than in the atmosphere, thus
providing more time to collect detailed experimental data. (See Problem 7.53.)
Flow characteristics for a 30-ft-diameter prototype parachute are to be determined by tests
of a 1-ft-diameter model parachute in a water tunnel. Some data collected with the model
parachute indicate a drag of 17 lb when the water velocity is 4 ft/s. Use the model data to
predict the drag on the prototype parachute falling through air at 10 ft/s. Assume the drag
to be a function of the velocity, V , the fluid density, ρ , and the parachute diameter, D.
Solution 7.53
= F
= f (V , ρ , D )
ρ = FL−4T 2
V = LT
−1
D = L
From the pi theorem, 4 − 3 = 1 pi terms required, and a dimensional analysis yields
ρV 2 D 2
=C
where C is a constant. Thus, for similarity between model and prototype
ρV D
2
2
=
m
ρ mVm 2 Dm 2
So that
2
 ρ  V   D 


 

 ρ m  Vm   Dm 
2
m

ft 
−3 slugs  
 2.38 × 10
3   10
ft 
s   30 ft  (17 lb )
=



 1.94 slugs
  4 ft   1ft 
ft 3

 s 
= 117 lb
Problem 7.54
When small particles of diameter d are transported by a moving fluid having a velocity V ,
they settle to the ground at some distance  after starting from a height h as shown in the
figure below. The variation in  with various factors is to be studied with a model having a
1
length scale of . Assume that
10
 = f ( h, d ,V , γ , μ )
where γ is the particle specific weight and μ is the fluid viscosity. The same fluid is to be
used in both the model and the prototype, but γ ( model ) = 9 × γ (prototipe). (a) If
V = 50 mph, at what velocity should the model tests be run? (b) During a certain model test,
it was found that ( model ) = 0.8 ft . What would be the predicted  for this test?
Particle
V
h
ℓ
Solution 7.54
 = f ( h, d ,V , γ , μ )
 = L
h = L
V = LT −1
d = L
γ = FL−3
μ = FL−2T
From the pi theorem, 6 − 3 = 3 pi terms required, and a dimensional analysis yields
 d μV 

=φ , 2 
h
 h γh 
(a) For similarity
μ mVm μV
=
γ m hm2 γ h2
so that
2
γ μ  hm 
Vm = m
V
γ μ m  h 
and for the data given
2
1
Vm = ( 9 )(1)   ( 50 mph ) = 4.50 mph
 10 
(b) Since the prediction equation is
 m
=
h hm
it follows for the data given
=
h
 m = (10 )( 0.8ft ) = 8.00 ft
hm
Problem 7.55
A thin layer of an incompressible fluid flows steadily over a horizontal smooth plate as
shown in the figure below. The fluid surface is open to the atmosphere, and an obstruction
having a square cross section is placed on the plate as shown. A model with a length scale
1
of and a fluid density scale of 1.0 is to be designed to predict the depth of fluid, γ , along
4
the plate. Assume that inertial, gravitational, surface tension, and viscous effects are all
important. What are the required viscosity and surface tension scales?
Free surface
y
V
Solution 7.55
A fluid dynamics problem for which inertial, gravitational, surface tension, and viscous
effects are all important requires Froude, Reynolds, and Weber number similarity.
Thus, for
Vm
V
(Froude number similarity)
=
gmd m
gd
it follows that (with g = gm)
Vm
dm
=
V
d
For Reynolds number similarity,
ρ mVm d m ρVd
=
μm
μ
and
μ m ρ m Vm d m ρ m
=
=
μ
ρ V d
ρ
3
dm dm ρm  dm  2
=
ρ  d 
d d
3
 1 2 1
= (1.0 )   = = 0.125
8
4
For Weber number similarity,
ρ mVm2 d m ρV 2 d
=
σm
σ
and
2
σ m ρ m Vm2 d m ρ m  d m  d m ρ m  d m 
=
=
=


σ
ρ V d
ρ  d  d
ρ  d 
2
1
1
= (1.0)   =
= 0.0625
 4  16
2
Problem 7.56
During a storm, a snow drift is formed behind a snow fence as shown in the figure below.
Assume that the height of the drift, h, is a function of the number of inches of snow
deposited by the storm, d; height of the fence, H; width of slats in the fence, b; wind speed,
V; acceleration of gravity, g; air density, ρ; and specific weight of snow, γs. (a) If this
problem is to be studied with a model, determine the similarity requirements for the model
and the relationship between the drift depth for model and prototype (prediction equation).
(b) A storm with winds of 30 mph deposits 16 in. of snow having a specific weight of
1
5.01 lb/ft3. A -sized scale model is to be used to investigate the effectiveness of a proposed
2
snow fence. If the air density is the same for the model and the storm, determine the
required specific weight for the model snow and required wind speed for the model.
b
b
Fence
Drift
V
H
H
h
Solution 7.56
(a)
hL
h = f (d , H , b,V , g, ρ , γ s )
d L
H L
V  LT −1 g  LT −2 ρ  FL−4T 2 γ s  FL−3
bL
From the pi theorem, 8 – 3 = 5 pi terms required, and a dimensional analysis yields
 d b ρg V 
h
=φ , ,
,

H H γ
H
gH 
s

(b) Thus, for similarity between model and prototype
ρm gm ρ g
=
and for ρm = ρ, gm = g
γ sm
γs
γ sm = γ = 5.00
lb
ft3
Also,
Vm
gmH
=
V
gH
Hm 1
=
H
2
Hm
1
Vm =
V =
(30 mph) = 21.2 mph
h
2
so that with gm = g and
Problem 7.57
Air bubbles discharge from the end of a submerged tube as shown in the figure below.
The bubble diameter, D, is assumed to be a function of the air flowrate, Q; the tube
diameter, d; the acceleration of gravity, g; the density of the liquid, ρ; and the surface
tension of the liquid, σ. (a) Determine a suitable set of dimensionless variables for this
problem. (b) Model tests are to be run on the Earth for a prototype that is to be operated
on a planet where the acceleration of gravity is 10 times greater than that on Earth. The
model and prototype are to use the same fluid, and the prototype tube diameter is 0.25 in.
Determine the tube diameter for the model and the required model flowrate if the
prototype flowrate is to be 0.001 ft3/s.
Q
d
D
Solution 7.57
(a)
DL
D = f (Q, d , g, ρ ,σ )
Q  L3T −1 d  L g  LT −2 ρ  FL−4T 2 σ  FL−1
From the pi theorem, 6 – 3 = 3 pi terms required, and a dimensional analysis yields
 σ
D
Q 
=φ
,
 ρ gd 2 gd 52 
d


(b) For similarity
σm
=
ρ m g m d m2
σ
ρ gd 2
and with σm = σ, ρm = ρ
dm =
10 g m
g
d=
(0.25 in.) = 0.791in.
gm
gm
Also,
Qm
5
2
gm dm
=
Q
5
gd 2
so that
5
3
gm  d m  2
1  0.791in.  2 
ft3 
−3 ft
Qm =
Q=

  0.001  = 5.63 × 10
10  0.25 in.  
s 
s
g  d 
With the above similarity requirements met
Dm D
Dm d m
or
=
=
= 10
d
d
D
d
5
Problem 7.58
For a certain model study involving a 1:5 scale model, it is known that Froude number
similarity must be maintained. The possibility of cavitation is also to be investigated, and it
is assumed that the cavitation number must be the same for model and prototype. The
prototype fluid is water at 30 °C, and the model fluid is water at 70 °C. If the prototype
operates at an ambient pressure of 101 kPa (abs), what is the required ambient pressure for
the model system?
Solution 7.58
For Froude number similarity,
Vm
gm  m
=
V
g
so that (with g = gm)
Vm
=
V
m

(1)
For cavitation number similarity,
( pr = pv )m
1 ρ V2
2 m m
=
( pr − pv )
1
ρV 2
2
It follows that
( pr − pv ) m =
ρ m Vm2
(p − p )
ρ V2 r v
( pr − pv )m =
ρm  m
(p − p )
ρ  r v
And making use of Eq. (1)
(2)
For water
@ 70 °C
ρ m = 977.8
@ 30 °C
ρ = 995.7
kg
m
kg
m
3
3
; pv m = 3.116 × 10 4
; pv = 4.243 × 103
N
m2
N
m2
( abs )
( abs )
Thus, from Eq. (2)
kg 

 977.8 3  1
m   101×103 N − 4.243 ×103 N  + 3.116 ×10 4 N
prm = 

kg   5 

m2
m2 
m2

995.7


m3 

= 50.2 kPa ( abs )
Problem 7.59
A model hydrofoil is to be tested. Is it practical to satisfy both the Reynolds number and
the Froude number for the hydrofoil when it is operating near the water surface? Support
your decision.
Solution 7.59
The Reynolds number requirements is
Vm  m Vp  p
=
vm
vp
The Froude number requirements is
Vm
g m
=
Vp
g p
.
Thus
Vm

= m.
Vp
p
Substituting into the Reynolds number requirements gives
3
vm Vm  m   m  2
=
=

v p V p  p   p 
Since the model is probably smaller than the prototype, the test must be conducted in a
liquid with kinematic viscosity significantly less than the prototype fluid (water). The only
practical test fluid is also water. It is possible to change the viscosity of water by heating; if
the prototype were to operate in 20 °C water, testing with 80 °C water would give a
2
viscosity ratio of 0.365. The scale factor could then be as low as (0.365) 3 = 0.51
Matching both Re and Fr is impractical.
Problem 7.60
A thin layer of particles rests on the bottom of a horizontal tube as shown in the figure
below. When an incompressible fluid flows through the tube, it is observed that at some
critical velocity, the particles will rise and be transported along the tube. A model is to be
used to determine this critical velocity. Assume the critical velocity, Vc, to be a function of
the pipe diameter, D, particle diameter, d, the fluid density, ρ, and viscosity, μ, the density
of the particles, ρp, and the acceleration of gravity, g. (a) Determine the similarity
requirements for the model, and the relationship between the critical velocity for model and
1
prototype (the prediction equation). (b) For a length scale of and a fluid density scale of
2
1.0, what will be the critical velocity scale (assuming all similarity requirements are
satisfied)?
Vc
Solution 7.60
Vc = f ( D, d , ρ , μ , ρ p , g )
(a)
Vc  LT −1
ρ p  FL−4T −2 g  LT −2
ρ  FL−4T 2 μ  FL−2T
d L
DL
From the pi theorem, 7 – 3 = 4 pi terms required, and a dimensional analysis yields
 d ρ gd 3 ρ 2 
ρVc D
=φ ,
,

 D ρp
μ
μ 2 

Thus, the similarity requirements are
ρm
ρ
=
ρ pm ρ p
dm d
=
Dm D
3 2
gmd m
ρm
μ m2
=
gd 3 ρ 2
μ2
The prediction equation is
ρVc D ρ mVcm Dm
=
μ
μm
(b) If all similarity requirements are satisfied, the prediction equation indicates that
Vcm
μ
μ 
ρ μm D
=
= (1.0)  m  (2) = 2 m
ρ m μ Dm
μ
Vc
 μ 
From the third similarity requirement (with g − gm),
3
2
3
1
μm
d  ρ 
1
=  m   m  =   (1.0)2 =
μ
8
2
 d   ρ 
Thus, from Eq. (1)
Vcm
1
=2
= 0.707
8
Vc
(1)
Problem 7.61
The pressure rise, ∆p, across a blast wave, as shown in the figure below is assumed to be a
function of the amount of energy released in the explosion, E; the air density, ρ; the speed
of sound, c; and the distance from the blast, d. (a) Put this relationship in dimensionless
form. (b) Consider two blasts: the prototype blast with energy release E and a model blast
with 1/1000 th the energy release (Em = 0.001E ). At what distance from the model blast
will the pressure rise be the same as that at a distance of 1 mile from the prototype blast?
Air (ρ , c)
Δp = p2 – p1
(2) (1)
d
Solution 7.61
Δp = f ( E , ρ , c, d , )
(a)
Δp  FL−2
ρ  FL−4T 2
E  FL
c  LT −1
d L
From the pi theorem, 5 – 3 = 2 pi terms required, and a dimensional analysis yields
 E 
Δp
=
φ

2 3
ρc2
 ρc d 
(b) For similarity,
Em
2 3
dm
ρ m cm
and with ρm = ρ, cm = c, it follows that
3
dm
=
For
Em
= 0.001 and d = 1 mi
E
=
E
ρc2d 3
Em 3
d
E
d m3 = (0.001)(1mi)3
d m = 0.100 mi
With this similarity requirement satisfied, the prediction equation is
Δpm
Δp
= 2
2
ρ m cm ρ c
and therefore Δpm = Δp
at
d m = 0.100 mi
Problem 7.62
An incompressible fluid oscillates harmonically (V = V0 sin(ω t), where V is the velocity)
1
with a frequency of 10 rad/s in a 4-in.-diameter pipe. A scale model is to be used to
4
determine the pressure difference per unit length, Δp (at any instant) along the pipe.
Assume that
Δp = f (D,V0 ,ω, t, μ, ρ )
where D is the pipe diameter, ω is the frequency, t is the time, μ is the fluid viscosity, and ρ
is the fluid density. (a) Determine the similarity requirements for the model and the
prediction equation for Δp . (b) If the same fluid is used in the model and the prototype, at
what frequency should the model operate?
Solution 7.62
Δp  FL−3
DL
V0  LT −1
ω  T −1
t T
μ  FL−2T
ρ  FL−4T 2
From the pi theorem, 7 – 3 = 4 pi terms required, and a dimensional analysis yields
DΔp
ρV02
ρV D 
V t
= φ  0 , ωt, 0 
μ 
 D
(a) Thus, the similarity requirements are
V0 mtm V0t
=
ωmtm = ωt
Dm
D
ρ mV0mDm ρV0 D
=
μm
μ
and the prediction equation is
DΔp
ρV02
=
Dm Δp m
ρ mV02m
(b) For Reynolds number similarity (the last similarity requirement), with the same fluid in
model and prototype,
V0 m
D
=
V0
Dm
So that from the first similarity requirement
tm Dm V0  Dm  Dm   Dm 
=
=

=

t
D V0 m  D  D   D 
2
Thus, to satisfy the remaining similarity requirements
ωmtm = ωt
or
2
 D 
t
rad 
rad
2
ωm = ω = 
 ω = (4)  10
 = 160
s 
s
tm

 Dm 
Problem 7.63
As shown in the figure below, a “noisemaker” B is towed behind a minesweeper A to set off
enemy acoustic mines such as at C. The drag force of the noisemaker is to be studied in a
1
water tunnel at a scale model (model 1/4 the size of the prototype). The drag force is
4
assumed to be a function of the speed of the ship, the density and viscosity of the fluid, and
the diameter of the noisemaker. (a) If the prototype towing speed is 3 m/s, determine the
water velocity in the tunnel for the model tests. (b) If the model tests of part (a) produced a
model drag of 900 N, determine the drag expected on the prototype.
A
B
C
Solution 7.63
(a)  = f (V , ρ , μ, D), where   F =
ML
T
2
,V 
L
M
M
and D  L
,ρ = 3 ,μ 
T
LT
L
Thus, k − r = 5 − 3 = 2 so that Π1 = φ ( Π 2 ),
where by inspection there are the ingredients for a Reynolds number, Re =
coefficient, CD =

1
2 2
 ρV D 
2

.
Hence,
CD = φ (Re)
For similarity, Rem = Re, or
ρ mVmDm ρVD
=
μm
so that with ρm = ρ and μm = μ,
μ
VmDm = VD
Vm =
or with Dm =
1
D,
4
D
m
 m
V = 4V = 4  3  = 12
s
Dm


(b) With Rem = Re, it follows that C D m = CD , or
m
1
ρ mVm2 Dm2
2
=

1
ρV 2 D2
2
ρVD
, and a drag
μ
Thus, since ρm = ρ,
m
=

or
V D2
2
 m 
2
2
3s 
V   D 
2
=

=
 
 m  m  (4) (900 N) = 900 N
 Vm   Dm 
 12 
s 

Note: The prototype has the same drag as the model.
2
Vm2 Dm
2
Problem 7.64
The drag characteristics for a newly designed automobile having a maximum characteristic
length of 20 ft are to be determined through a model study. The characteristics at both low
speed (approximately 20 mph) and high speed (90 mph) are of interest. For a series of
projected model tests, an unpressurized wind tunnel that will accommodate a model with a
maximum characteristic length of 4 ft is to be used. Determine the range of air velocities
that would be required for the wind tunnel if Reynolds number similarity is desired. Are the
velocities suitable? Explain.
Solution 7.64
For Reynolds number similarity,
ρ mVm  m ρV 
=
μm
μ
so that
Vm =
μm ρ 
V
μ ρm  m
(1)
Since the wind tunnel is unpressurized, the air properties will be approximately the same for
model and prototype. Thus, Eq. (1) reduces to

Vm =
V
m
and for the data given
Vm =
(20 ft)
20 ft
V = 5V Vm =
=5
(44 ft)
4 ft
Therefore, at low speed,
Vm = 5(20 mph) = 100 mph
and at high speed,
Vm = 5(90 mph) = 450 mph
So that the model velocity range is 100 mph to 450 mph .
At the high wind tunnel velocity, compressibility of air would start to become an important
factor, whereas compressibility is not important for the prototype. Thus, the higher velocity
required for the model would not be suitable.
No.
Problem 7.65
The drag characteristics of an airplane are to be determined by model tests in a wind tunnel
operated at an absolute pressure of 1300 kPa. If the prototype is to cruise in standard air at
385 km/hr, and the corresponding speed of the model is not to differ by more than 20%
from this (so that compressibility effects may be ignored), what range of length scales may
be used if Reynolds number similarity is to be maintained? Assume the viscosity of air is
unaffected by pressure, and the temperature of air in the tunnel is equal to the temperature
of the air in which the airplane will fly.
Solution 7.65
For Reynolds number similarity,
ρ mVm  m ρV 
=
μm
μ
so that
 m μm ρ V
=
μ ρ m Vm

For an ideal gas, p = ρRT, and with constant temperature
p
= constant
ρ
or
p
ρ
=
pm ρ m
and Eq. (1) can be written as (with μm = μ)
p V
m
=
pm Vm

For the data given
m
(101kPa ) V
=

(1300 kPa ) Vm
and with Vm = (1 ± 0.2 )V , it follows that
 m (101kPa )
1
=

(1300 kPa ) (1 ± 0.2 )
Thus, the range of length scales is 0.0647 to 0.0971.
(1)
Problem 7.66
The drag on a sphere moving in a fluid is known to be a function of the sphere diameter,
the velocity, and the fluid viscosity and density. Laboratory tests on a 4-in.- diameter
sphere were performed in a water tunnel and some model data are plotted in the figure
below. For these tests, the viscosity of the water was 2.3 × 10−5 lb ∙ s/ft2 and the water
density was 1.94 slugs/ft3. Estimate the drag on an 8-ft- diameter balloon moving in air at a
velocity of 3 ft/s. Assume the air to have a viscosity of 3.7 × 10−7 lb ∙ s/ft2 and a density of
2.38 × 10−3 slugs/ft3.
Model drag (lb)
6
4
2
0
0
2
4
6
8
10
12
Model velocity (ft/s)
Solution 7.66
 = f (d ,V , ρ , μ )
where:  ~ drag  F , d ~ sphere diameter  L, V ~ velocity  LT −1,
ρ ~ fluid density  FL−4T 2, μ ~ fluid viscosity  FL−2T .
From the pi theorem, 5 – 3 = 2 pi terms required, and a dimensional analysis yields
 ρVd 
=φ

ρV d
 μ 

2 2
Thus, Reynolds number similarity is required so that
ρ mVm d m ρVd
=
μm
μ
or
Vm
=
μm ρ d
V
μ ρm dm
slugs 

−5 lb ⋅ s  

2.38 ×10 −3 3  
 2.3 ×10
2 

  ft 
8ft
ft
ft  
ft 
= 
 4   3  = 5.49
slugs 
s


 s
−7 lb ⋅ s 
 3.7 ×10
1.94 3   12 ft 
2 
ft 
ft 


From the graph, for Vm = 5.49
ft
, m = 1.30 lb. Since
s

ρV d
2 2
=
m
ρ mVm2 d m2
or
=
ρ V 2 d2

ρ m Vm2 d m2 m
so that
2

−3 slugs   ft 
3
 2.38 ×10
3   s 
(8ft)2
ft




(1.30 lb) = 0.274 lb
=
2
2
slugs 

ft
4

 

1.94 3 
 5.49   ft 
ft 

s
12

 

Problem 7.67
A dam spillway is 40 ft long and has fluid velocity of 10 ft/s. Considering Weber number
effects as minor, calculate the corresponding model fluid velocity for a model length of 5 ft.
Solution 7.67
The primary similarity parameter is Froude number,
Vm
gLm
=
Vp
gLp
.
Then
Vm = V p
Lm
ft
and Vm = 10
s
Lp
5
ft
= 3.54
40
s
Problem 7.68
A flagpole has a diameter of 1.0 ft and is 45 ft. long. The manufacturer wants to find the
bending moment at the base of the pole for a wind speed of 60 mph. To do so, a scale
model of the flagpole is built with a diameter of 1.0 in. and a length of 45 in. (a) What must
be the air speed at which the model is tested? (b) If the wind velocity over the flagpole in
part (a) is reduced by 50%, how much would the bending moment at the base be reduced?
Solution 7.68
d
We first develop the dimensionless groups and will use the pi theorem.
Identify the relevant dimensional parameters,
L
V = wind velocity  
T 
ℓ = length (L)
V
ℓ
d = diameter (L)
M = bending moment (FL)
I = second moment of area (L4)
m
F 
E = young’s modulus  2 
L 
 M FT 2 
ρ = air mass density  3 = 4 
L 
L
Determine the number of dimensionless parameters. There are three fundamentals
(F, L, T). Since E, V, and d cannot form a dimensionless group, k = 3 and n – k = 7 – 3 = 4
dimensionless groups.
Select E, V, and d as repeating parameters. Two obvious dimensionless groups are
I
and Π 2 = 4
Π1 =
d
d
We now combine E, V, and d with M
[M][ E ]a [V ]b [ d ]c = F 0 L0T 0
or
a
b
 F  L
( FL )  2    ( L )c = F 0 L0T 0 .
 L  T 
Equating power of F, L, and T gives
F:
L:
T:
1+ a = 0
 a = −1

1 − 2a + b + c = 0  b = 0
 c = 2a − 1 − b = −3
−b = 0

Then
Π3 =
M
E 3
We now combine E, V, and d with ρ.
[ ρ ][ E ]a [V ]b [d ]c = F 0 L0T 0
or
a
b
 1  F   L 
c
0 0 0
 4  2    ( L ) = F L T .
T
 L  L   
Equating power of F, L, and T gives
F:
L:
T:
1+ a = 0
 a = −1

−4 − 2a + b + c = 0  b = 2
 c = 4 + 2a − b = 0
2−b = 0

Then
Π4 =
ρV 2
E
Equating Π4 for the model and prototype gives
ρ mVm2 ρ pV p
.
=
Em
Ep
2
Equating Π3 for the model and prototype gives
Mp
Mm
=
E 3m E 3p
Eliminating Em and Ep gives
ρ mVm2 3m ρ pV p  p
=
Mm
Mp
2 3
or
Vm = V p
ρ pMm 3p
ρ mM p 3m
If we assumed ρm = ρp (same air temperature),
Vm = V p
M m 3p
3
M m  12 
= 60 mph
 
3
Mp  1 
M p m
or
.Vm = 2490 mph
Mp
Mm
Choosing
3
Mm   m 
=

M p   p 
gives
Vm = V p = 60 mph
Problem 7.69
1
A very small needle valve is used to control the flow of air in a -in. air line. The valve has a
8
pressure drop of 4.0 psi at a flowrate of 0.005 ft 3 s of 60 F air. Tests are performed on a
large, geometrically similar valve and the results are used to predict the performance of the
smaller valve. How many times larger can the model valve be if 60 F water is used in the
test and the water flowrate is limited to 7.0 gal min ?
Solution 7.69
We will match the Reynolds Number
Re =
The velocity is
V=
Vpvm
D
ρVD VD VmDm Vp Dp
=

=
 m=
μ
v
vm
vp
Dp Vmv p
Q
Q
=
A π D2
4
So
Qp
v
2 m
D
D p Q p vm
Qmv p
Dm
D
p
=
=

=
 m=
Dp Vmv p Qm v
Dm Qmv p
D p Q p vm
2 p
Dm
Vpvm
From property tables vm = v60 oF water = 1.22 ×10 −5
ft 2
s
v p = v60 o F air = 1.57 × 10 −4
ft 2
s
ft 2
gal
1.57 ×10 −4
Dm
s = 40
min
=
3
Dp
ft
gal
s
ft 2
0.005 × 7.48 3 × 60
1.22 ×10 −5
s
min
s
ft
7.0
So
Comment When using the Property Tables, the kinematic viscosity of air was evaluated at
atmospheric pressure. Actually, the pressure of the air is unknown. If it is near atmospheric,
the 4.0 psi pressure drop would cause compressibility effects and Mach Number must also
be considered.
Problem 7.70
In low-speed external flow over a bluff object, vortices are shed from the object (as shown
in the figure below). The frequency of vortex shedding, f ≐ 1/T, depends on ρ, μ, V, and d.
Make a dimensional analysis of this phenomenon. Transform the dimensionless parameters
that you developed into “standard” parameters. What are their names? Two identically
shaped objects with a size ratio 2:1 are tested in air flow. What velocity ratio is necessary
for dynamic similarity? What is the ratio of vortex-shedding frequencies?
V
d
Solution 7.70
The dimensional parameters have been identified. Their dimensions are
1
f
T
M
ρ
L3
F ⋅T
M
μ
=
2
( L ⋅T )
L
L
V
T
d
L
The number of dimensionless parameters will be n – k = 5 – 3 = 2. Note that k = 3 (the
number of fundamental dimensions) since ρ, V, and d cannot form a dimensionless group.
Select the repeating parameters as ρ, V, and d. We now combine ρ, V, and D with f to get
[ f ][ ρ ]a [V ]b [ d ]c = M 0 L0T 0 ,
 1  M
   3
 T  L
a
b
L
0 0 0
c
   ( L ) = M L T ,
T
 
M:
a=0
 a=0

L : −3a + b + c = 0  b = −1
− 1 − b = 0  c = 3a − b = 1
T:
The first dimensionless parameter is
Π1 = f ρ 0V −1d or
Π1 =
fd
V
Next combine ρ V, and D with μ to get
[ μ ][ ρ ]a [V ]b [ d ]c = M 0 L0T 0 ,
a
b
 M  M   L 
c
0 0 0

  3    (L) = M L T ,
 LT   L   T 
1 + a = 0  a = −1
M:

L : −1 − 3a + b + c = 0  b = −1
− 1 − b = 0  c = 1 + 3a − b = −1
T:
The second dimensionless parameter is
Π2 = μρ −1V −1d −1 or Π2 =
μ
ρVd
The standard dimensionless parameters are
Π1 = St (Strouhal Number)
Π 2−1 = Re (Reynolds Number)
Equality of the Reynolds number gives
 ρVd   ρVd 

 =

 μ 1  μ 2
For (assumed) equal air temperature and pressure, ρ1 = ρ2 and μ1 = μ2 so
V1 d2
=
=2
V2 d1
or
1 → smaller d
V1
= 2,
2 → larger d
V2
Equality of the Strouhal number gives
 fd   fd 
  =  ,
 V 1  V 2
f1  2 
=   (2), or
f2  1 
f1  d2  V1 
=    ,
f2  d1   V2 
1 → smaller d
f1
= 4,
2 → larger d
f2
Problem 7.71
In a nutritional products plant, a plate heat exchanger cools infant formula by
approximately 25 °C from an inlet temperature of 68 °C as it flows through the plates as
shown in the figure below. The process engineer would like to be able to cool the infant
formula by as much as 35 °C as it passes through the plates. The plate heat exchanger is
1
currently in use, so he would like to build a -scale model for testing purposes. In the
2
model, water will be used instead of infant formula. What flowrate is needed in the model?
D = 0.05 m
Infant formula
ρ = 1036.6 kg/m3
μ = 1.7 × 10−3 N·s/m2
Q = 16.8 m3/hr
at average formula
temperature
1.4 m
30% glycol solution
Solution 7.71
The Reynolds number must be the same for the model (water) and the prototype (formula).
 ρV  
 ρV  

 =

 μ m  μ  p
Since Q =
π
4
D2V
  4Q   
  4Q   
= ρ 
ρ 
2 μ
2 
  π D  m   π D  μ  p
or
2
  p   Dm   ρ p   μm 
Qm = Q p 
 
.


  m   Dp   ρ m   μ p 
Using an average temperature of 50 °C water,
ρ m = 988
kg
m
3
and μ m = 5.47 × 10 −4
3

m2
. Also Dm = 0.5Dp and m = 0.5 p. Then
kg  
N ⋅s 
5.47 × 10 −4 2 
3 
m 
m 
kg  
−3 N ⋅ s 
988 3   1.7 ×10

m 
m2 
2  1036.6

m  2  1 
Qm = 16.8
   

hr   1   2  



m3
Qm = 2.84
hr
N ⋅s
Problem 7.72
The pressure rise, p, across a centrifugal pump of a given shape can be expressed as
Δp = f ( D,ω, ρ,Q)
where D is the impeller diameter, ω is the angular velocity of the impeller, is the fluid
density, and Q is the volume rate of flow through the pump. A model pump having a
diameter of 8 in. is tested in the laboratory using water. When operated at an angular
velocity of 40π rad/s the model pressure rise as a function of Q is shown in the figure below.
Use this curve to predict the pressure rise across a geometrically similar pump (prototype)
for a prototype flowrate of 6 ft3/s. The prototype has a diameter of 12 in. and operates at an
angular velocity of 60π rad/s. The prototype fluid is also water.
8
Model data
ωm = 40π rad/s
Dm = 8 in.
Δ pm (psi)
6
4
2
0
0
0.5
1.0
1.5
2.0
Qm (ft3/s)
Solution 7.72
Δp = f ( D, ω, ρ ,Q )
−2
Δp  FL
D  L ω  T −1
ρ  FL−4T 2 Q = L3T −1
From the pi theorem, 5 – 3 = 2 pi terms required, and a dimensional analysis yields
Δp
 Q 
=φ

ρω D
 ω D3 
2
2
Thus, the similarity requirement is
Qm
Q
=
3
ωm Dm ω D3
so that
3
 ω  D 
Qm =  m  m  Q
 ω  D 
and for the data given
rad 

3
 40π

3
ft3
s   8 in   ft 

Qm =
=
6
1.19


rad   12 in   s 
s

π
60


s 

ft3
From the graph, pm = 5.50 psi for Q = 1.19 . Thus,
s
Δp
ρω D
2
and with =
2
=
Δpm
ρ mω m2 Dm2
m
2
rad 

60π
2


 ω   D 
s   12 in 

Δp = 
( 5.50 psi )
 
 Δp =
2  8 in 
rad


 ωm   Dm 


 40π

s


= 27.8psi
2
2
Problem 7.73
At a large fish hatchery, the fishes are reared in open, water-filled tanks. Each tank is
approximately square in shape with curved corners, and the walls are smooth. To create
motion in the tanks, water is supplied through a pipe at the edge of the tank. The water is
drained from the tank through an opening at the center. A model with a length scale of 1:13
is to be used to determine the velocity, V, at various locations within the tank. Assume that
V = f( , i, , μ, g, Q) where is some characteristic length such as the tank width, i
represents a series of other pertinent lengths, such as inlet pipe diameter, fluid depth, etc.,
is the fluid density, μ is the fluid viscosity, g is the acceleration of gravity, and Q is the
discharge through the tank. (a) Determine a suitable set of dimensionless parameters for
this problem and the prediction equation for the velocity. If water is to be used for the
model, can all of the similarity requirements be satisfied? Explain and support your answer
with the necessary calculations. (b) If the flowrate into the full-sized tank is 250 gpm,
determine the required value for the model discharge assuming Froude number similarity.
What model depth will correspond to a depth of 32 in. in the full-sized tank?
Solution 7.73
(a)
V = f ( , i , ρ , μ , g ,Q )
From the pi theorem, 7 – 3 = 4 pi terms required, and a dimensional analysis yields

V 2
Q 2 ρQ 
=φ i , 5 ,


μ
Q
g


Thus, the similarity requirements are
 im  i
=
m

2
Qm
5m g m
=
ρ mQm ρQ
=
 m μ m μ
Q2
5
 g
and the prediction equation is
V  2 Vm  2m
=
Q
Qm
From the last similarity requirement with
m
= and μm = μ
Qm
ρ μm  m  m
=
=
ρm μ 

Q
However, from the second similarity requirements with gm = g
5
Qm   m  2
=
Q   
Since these two requirements are in conflict it follows that the similarity requirements
cannot be satisfied. No.
(b) For Froude number similarity
Vm
V
=
gm m
g
and with gm = g
Vm
m
=
.
V

Thus, from the prediction equation
V  2 Vm  2m
=
Q
Qm
It follows that
2
2
5
Qm Vm   m 
 m   m    m 2
=
=
=


Q
V   
      
So that with
m 1
=
13

5
 1 2
Qm =   (250 gpm) = 0.410 gpm
 13 
Note that this same result can be obtained from the second similarity requirements (which
corresponds to Froude number similarity) since
2
Qm
5m g m
=
Q2
5 g
And therefore
5
  2
Qm =  m  Q
  
Geometric similarity requires that
 im  i
=
m

or
 im  m 1
=
=
i
 13
so that all lengths scale as the length scale. Thus,
( depth )model = 
1
 ( depth )prototype
 13 
1
=   ( 32 in.) = 2.46 in.
 13 
Problem 7.74
The Tacoma Narrows Bridge failure is a dramatic example of the possible serious effects of
wind-induced vibrations. As a fluid flows around a body, vortices may be created that are
shed periodically, creating an oscillating force on the body. If the frequency of the shedding
vortices coincides with the natural frequency of the body, large displacements of the body
can be induced as was the case with the Tacoma Narrows Bridge. To illustrate this type of
phenomenon, consider fluid flow past a circular cylinder. Assume the frequency, n, of the
shedding vortices behind the cylinder is a function of the cylinder diameter, D; the fluid
velocity, V; and the fluid kinematic viscosity, v. (a) Determine a suitable set of
dimensionless variables for this problem. One of the dimensionless variables should be the
Strouhal number, nD/V. (b) Some results of experiments in which the shedding frequency of
the vortices (in Hz) was measured, using a particular cylinder and Newtonian,
incompressible fluid, are shown in the figure below. Is this a “universal curve” that can be
used to predict the shedding frequency for any cylinder placed in any fluid? Explain. (c) A
certain structural component in the form of a 1-in.-diameter, 12-ft-long rod acts as a
cantilever beam with a natural frequency of 19 Hz. Based on the data in the figure below,
estimate the wind speed that may cause the rod to oscillate at its natural frequency.
Hint: Use a trial-and-error solution.
0.22
St = nD/ V
0.20
0.18
0.16
0.14
0.12
10
100
1,000
10,000
Re = VD/v
Solution 7.74
(a)
n = f ( D,V , v )
n  T −1 D  L V  LT −1 v  L2T −1
From the pi theorem, 4 – 2 = 2 pi terms required, and a dimensional analysis yields
nD
 VD 
=φ

V
 v 
(b) Yes. If the variable of part (a) are correct then this is a “universal” or general
relationship between the Strouhal number and the Reynolds number. It is valid over the
range of Reynolds number covered in the experiment.
1in.
1
= ft
in. 12
12
ft
1 
19Hz )  ft 
(
nD
 12 
=
St =
V
V
From the figure in the problem, assume St = 0.21 and from Eq. (1)
(c) For n = 19 Hz and D =
0.21 =
so that
(1)
(19Hz ) 
V = 7.54
1 
ft 
 12 
V
ft
s
(5.14 mph)
ft   1 

7.54   ft 

VD 
s   12 
Check Re : Re =
=
= 4000
2
v
−4 ft
1.57 × 10
s
From the figure in the problem, at Re = 4000, St = 0.21, and therefore, assumed value of
St OK.
Problem 7.75
A model study is to be developed to determine the force exerted on bridge piers due to
floating chunks of ice in a river. The piers of interest have square cross sections. Assume
that the force, R, is a function of the pier width, b; the thickness of the ice, d; the velocity of
the ice, V; the acceleration of gravity, g; the density of the ice, ρi; and a measure of the
strength of the ice, Ei, where Ei has the dimensions FL−2. (a) Based on these variables
determine a suitable set of dimensionless variables forth is problem. (b) The prototype
conditions of interest include an ice thickness of 12 in. and an ice velocity of 6 ft/s. What
model ice thickness and velocity would be required if the length scale is to be 1/10? (c) If the
model and prototype ice have the same density, can the model ice have the same strength
properties as that of the prototype ice? Explain.
Solution 7.75
(a)
R = f ( b, d ,V , g, ρi , Ei )
RF
b  L d  L V  LT −1 g  LT −2
ρi  FL−4T 2 Ei  FL−2
From the pi theorem, 7 – 3 = 4 pi terms required, and a dimensional analysis yields
 b V 2 ρiV 2 
R
=
,
,
φ

 d gd E 
Ei b2
i 

(b) For similarity,
bm b
=
dm d
or
d m bm 1
=
=
d
b 10
so that
dm =
1
(12 in.) = 1.20 in.
10
Also,
Vm2
V2
=
g m d m gd
(1)
and with gm = g
 g  d 
ft
 1   ft 
Vm =  m   m V = (1)    6  = 1.90
s
 10   s 
 g  d 
(c) For similarity,
ρimVm2
Eim
=
ρV 2
Ei
Thus,
2
Eim  ρim   Vm 
d
=
= m


Ei  ρi   V 
d
since ρim = ρi and
from Eq.(1)
2
dm
 Vm 
V  = d


dm 1
Since
= , Eim ≠ Ei and model ice cannot have same strength properties as that of the
d
10
prototype ice. No.
Problem 7.76
Models are commonly used to study the dispersion of a gaseous pollutant from an exhaust
stack located near a building complex. Similarity requirements for the pollutant source
involve the following independent variables: the stack gas speed, V; the wind speed, U; the
density of the atmospheric air, ; the difference in densities between the air and the stack
gas, − s; the acceleration of gravity, g; the kinematic viscosity of the stack gas, vs; and the
stack diameter, D. (a) Based on these variables, determine a suitable set of similarity
requirements for modeling the pollutant source. (b) For this type of model, a typical length
scale might be 1:200. If the same fluids were used in model and prototype, would the
similarity requirements be satisfied? Explain and support your answer with the necessary
calculations.
Solution 7.76
(a) Since
V  LT −1 U  LT −1 ρ  FL−4T 2
ρ − ρs  LT −4T 2 g  LT −2 vs  L2T −1 D  L,
it follows from the pi theorem, 7 – 3 = 4 pi terms are required. A dimensional analysis yields
ρ − ρs
V VD V 2
,
,
and
ρ
U vs gD
as a possible set of pi terms. Thus, the similarity requirements would be:
Vm V
=
Um U
(b) For
Vm Dm VD
=
vsm
vs
Vm2
V2
=
g m Dm gD
( ρ − ρs )m ( ρ − ρs )
=
ρm
ρ
Dm
1
and vsm = vs the second similarity requirement is
=
D 200
Vm vsm D
=
= 200
(see above)
V
vs Dm
However, from the third similarity requirement with gm = g
Vm
Dm
1
=
=
200
V
D
This result conflicts with that from the second similarity requirement, and therefore the
similarity requirements cannot be satisfied under the stated conditions. No.
Problem 7.77
As winds blow past buildings, complex flow patterns can develop due to various factors
such as flow separation and interactions between adjacent buildings. Assume that the local
gage pressure, p, at a particular location on a building is a function of the air density, ρ; the
wind speed, V; some characteristic length, ; and all other pertinent lengths, i, needed to
characterize the geometry of the building or building complex. (a) Determine a suitable set
of dimensionless parameters that can be used to study the pressure distribution. (b) An
eight-story building that is 100 ft tall is to be modeled in a wind tunnel. If a length scale of
1:300 is to be used, how tall should the model building be? (c) How will a measured
pressure in the model be related to the corresponding prototype pressure? Assume the same
air density in model and prototype. Based on the assumed variables, does the model wind
speed have to be equal to the prototype wind speed? Explain.
Solution 7.77
(a)
−2
p = FL
p = f ( ρ ,V , ,  i )
ρi  FL−4T 2 V  LT −1   L  i  L
From the pi theorem, 5 − 3 = 2 pi terms required, and a dimensional analysis yields
  
=
φ
 
ρV 2
 i 
p
(b) For geometric similarity
m

=
 im  i
so that
 m  im
=

i
and it follows that all pertinent lengths are scaled with the length scale
m
1
100 ft
=
model height =
= 0.333 ft.
300
300

(c) With geometric similarity satisfied it follows that
pm
p
=
2
ρV
ρ mVm 2
m
. Thus, with

Thus, with ρm = ρ
2
V 
p=
 pm
 Vm 
Vm
, so the
V
model wind speed does not have to be equal to the prototype wind speed. No.
With the set of given variables there is no requirement for the velocity scale,
Problem 7.78
Assume that the pump performance characteristics shown in the figure below are for a
pump running at 1200 rpm. Assuming that the pump head, hp; the pump impeller diameter,
D; and the flowrate, Q, through the pump are the other relevant parameters, calculate the
pump performance characteristics at speeds of 1800 rpm and 2400 rpm. Neglect viscous
effects.
200
T = 55°F
Head (ft)
150
hp
100
50
0
0
25
50
75
100
125
150
Volume flowrate, Q (gal/min)
Solution 7.78
The dimensionless performance curve for a pump is
 Q ND 2 
gh
=
F
,


3
N 2D2
v 
 ND
For negligible Reynolds number effects,
 Q 
=F

N D
 ND3 
We can transform a curve at speed N1 to a curve at speed N2 by
gh
2
gh1
N12 D 2
=
2
gh2
and
N 22 D 2
Q1
N1D
3
=
Q2
N 2 D3
or
h2  N 2 
=

h1  N1 
2
and
Q2 N 2
=
Q1 N1
For our case, N1 = 1200 rpm and N2 = 1800 rpm and 2400 rpm. The performance curve at
1800 rpm is obtained by picking points (h,Q) from the curve at 1200 rpm and determining
the corresponding points at 1800 rpm from
2
 1800 
h1800 = 
 h1200 = 2.25h1200
 1200 
and
 1800 
Q1800 = 
 Q1200 = 1.5Q1200 .
 1200 
Similarity, the curve at 2400 rpm is found from
2
 2400 
h2400 = 
 h1200 = 4h1200
 1200 
and
 2400 
Q2400 = 
 Q1200 = 2Q1200 .
 1200 
Curves are shown in the following figure.
800
700
240
0
h (ft)
600
rpm
500
180
0 rp
m
400
300
1200 r
pm
200
100
20
40
60
80
100 120 140 160 180 200 220
Q (gpm)
Problem 7.79
A stream of atmospheric air is used to keep a ping-pong ball aloft by blowing the air
upward over the ball. The ping-pong ball has a mass of 2.5 g and a diameter D1 = 3.8 cm,
and the air stream has an upward velocity of V1 = 0.942 m/s. This system is to be modeled
by pumping water upward with a velocity V2 over a solid ball of diameter D2 and density
ρb 2 = 2710 kg/m3. In both cases, the net weight of the ball Wb is equal to the air drag,
C D ρ AV 2
,
2
where CD = 0.60, ρ is the fluid density, A the ball’s projected area, and V the velocity of the
fluid upstream from the ball. Determine all possible combinations of V2 and D2. [Hint: A
force balance involving the drag on the ball, the buoyant force on the ball, and the weight
of the ball is needed.]
Wb =
Solution 7.79
Both the ping-pong ball-air system and the other ball-water system must have the same
Reynolds number.
v 
or V2 D2 = V1D1  2  .
 v1 
Using kinematic viscosity values from tables of properties of air and water
V2 D2 V1D1
=
v2
v1

m2 
1.3 × 10 −6
2


m

s  = 0.00310 m .
V2 D2 =  0.942  ( 0.038 m ) 
2
s 
s


−5 m 
 1.5 ×10

s 

Since the ball’s net weight must equal the drag,
(
ρb − ρ2
2
)
gπ D23 C D ρ2  π
 2
=
 D2 V2
6
2 4

or
( 0.60 ) 1000
kg 

m3 
3 C D ρ2
3
s2

=
= 0.0268 .
m
kg
4
m
V2 4 ρb2 − ρ2
 9.81 2  ( 2710 − 1000 ) 3
s 
m

Solving the second equation for D2 and substituting this into the first equation gives
D2
=
2
(
)


s2
m2
V2 =  0.068 V22  = 0.00310


m
s


The diameter D2 is one solution
D2 =
or
V2 = 0.487
m2
m2
0.00310
s =
s = D = 0.00637 m
2
m
V2
0.487
s
0.00310
m
s
Both the ping-pong ball-air system and the other ball-water system must have the same
Reynolds number or
 D  v 
or V2 = V1  1  2  .
 D2  v1 
Using kinematic viscosity values from tables of properties of air and water,
V2 D2 V1D1
=
v2
v1
2

−5 m 
1.5
10
×


m   3.8cm 

s  = V = 6.94 m .

V2 =  1.90  

2
2
s   12.0 cm  
s

−6 m 
 1.3 × 10

s 

The net weight of the ball is the weight minus the buoyant force.
(
C D ρ2 A2V22
ρb2 − ρ2 gV2 =
,
2
)
π D23
( ρb − ρ2 ) g 6
2
C D ρ2π D22V22
=
,
8
kg  
m

3 ( 0.60 ) 1000 3   6.94 
2
3C ρ V
kg
s 

m 
ρb2 = ρ2 + D 2 2 = 1000 3 +
m
4 gD2

m
4  9.81 2  ( 0.12 m )
s 

ρb = 19400
2
kg
m3
Checking the densities of solids, a possible is
gold with ρb2 ≅ 19300
kg
m3
2
Problem 7.80
1
A -scale model of an airplane is tested in a wind tunnel at 70 °F and 14.40 psia. The
10
model test results are:
0
0
Velocity (mph)
Drag (lb)
50
5
100
21
150
46
200
85
Find the corresponding airplane velocities and drags if only fluid compressibility is
important and the airplane is flying in the U.S. Standard Atmosphere at 30,000 ft. Assume
that the air is an ideal gas.
Solution 7.80
If only compressibility is important, the Mach Numbers (or Cauchy Numbers) must be
matched:
Vp
Vp
V
Vm
Ma m = Ma p  m =

=
cm c p
kRTm
kRTp
R is a constant and at the temperatures involved k is nearly constant so
Vp
Vm
Tp
=
Tm
=
411.8 R
(460 + 70) R
 V p = 0.882Vm
(1)
If Mach Numbers are equal, then Drag Coefficients are equal
ρ p V p2  2p
C D ,m = C D , p 
=

=
1
1
Dm ρ m Vm2  2m
ρ V 2 2
ρ V 22
Dp
Dm
2
m m m
2
p p
Dp
p
Treating the air as an ideal gas
Dp
Dm
=
p p RTm Tp  2p
pmRTp Tm  2m
=
p p  2p
pm  2m
=
4.37 psia
(10)2  30.37Dm
14.4 psia
(2)
Using Eqs. (1) and (2) together with the given data:
V (mph)
D (lb)
0
0
44.1
151.8
88.2
637.8
132.3
1397
176.4
2581
Comment Because the top speeds are fairly low (200 mph/176 mph) compressibility effects
are fairly small and it might be better to match the Reynolds Number.
Problem 7.81
A company manufactures geometrically similar airplane propellers up to 4.0 m in diameter.
Wind tunnel tests are run on geometrically similar propellers up to 0.33 m in diameter. In
the test of a 0.33-m model of a 4.0-m propeller, the air velocity was 50 m/s. The model
rotational speed was 2000 rpm, the thrust 100 N, and the propeller input torque 20 N · m.
Calculate the corresponding prototype rotational speed, thrust, and input torque for an
airplane speed of 100 m/s. The Reynolds numbers and Mach numbers are of minor
importance.
Solution 7.81
First, we need to find appropriate dimensionless parameters for the propeller.
The primary parameters are speed ratio:
V
V
(A)
=
U ND
The “dependent” variable dimensionless parameters must involve thrust F and torque J .
Since thrust is a force (similar to lift or drag), we use a thrust coefficient
F
F
~
2 2
ρU D
ρ N 2 D5
Since torque equals force times length, we also use a torque coefficient
J
J
~
2 2
ρU D
ρ N 2 D5
The similarity requirement in equation (A) gives
Vp
Fp
Jp
Vm
Fm
Jm
=
,(B);
=
,(C);
=
,(D).
5
N p Dp N m Dm
ρ p N p2 Dp4 ρ m N m2 Dm4
ρ p N p2 D5p ρ m N m2 Dm
From (A)
m

 100 s   0.33 m 
=
=

 = 0.165
N m Vm Dp  50 m   4.0 m 
s 

Np
V p Dm
or
N p = 0.165 ( 2000 rpm ) = 330 rpm
From (B), noting
m
=
p
gives
2
4
4
 N p   Dp 
2  4.0 m 
=
 
 = ( 0.165 ) 
 = 587.7
Fm  N m   Dm 
 0.33 m 
Fp
or
Fp = ( 587.7 )(100 N ) = 58770 N
From (C), again noting
2
m
=
p
5
 N p   Dp   Fp   Dp 
 4.0 m 
=
 
 =

 ( 587.7 ) 
 = 7124
J m  N m   Dm   Fm   Dm 
 0.33 m 
Jp
or
J p = ( 7124 )( 20 N ⋅ m ) = 142000 N ⋅ m.
Problem 7.82
A breakwater is a wall built around a harbor so the incoming waves dissipate their energy
against it. The significant dimensionless parameters are the Froude number F and the
Reynolds number Re. A particular breakwater measuring 450 m long and 20 m deep is hit
1
-scale model of the breakwater, the
by waves 5 m high and velocities up to 30 m/s. In a
100
wave height and velocity can be controlled. Can complete similarity be obtained using
water for the model test?
Solution 7.82
The Reynolds and Froude number requirements are
Vp
Vm  m Vp  p
Vm
=
and
=
vm
vp
g m
g p
Assuming that vm = vp, the Reynolds number requirement is
Vm  p 1
=
=
Vp  m S
while the Froude number requirement is
Vm
m
=
= S
p
Vp
Obviously, these cannot both be true unless S = 1.
It is not possible to match both
Re and Fr in a scale model test.
Problem 7.83
A viscous fluid is contained between wide, parallel plates spaced a distance h apart as
shown in the figure below. The upper plate is fixed, and the bottom plate oscillates
harmonically with a velocity amplitude U and frequency ω. The differential equation for
the velocity distribution between the plates is
∂u
∂2u
ρ =μ 2
∂t
∂y
where u is the velocity, t is time, and ρ and μ are fluid density and viscosity, respectively.
Rewrite this equation in a suitable nondimensional form using h, U, and ω as reference
parameters.
Fixed plate
y
u
h
x
u = Ucos ω t
Solution 7.83
y
u
Let y* = , u* = , and t* = ωt so that:
h
U
( * ) ∂t* = U ∂u* (ω ) = Uω ∂u*
∂u ∂ Uu
=
∂t
∂t*
∂t
∂t*
∂u ∂ Uu
=
∂y
∂y*
∂y
 
∂y*  h 
∂t*
( * ) ∂y* = U ∂u*  1  = U ∂u*
∂2u
∂y
=
2
h ∂y*
U ∂  ∂u*  ∂y* U ∂ 2 u*
=


h ∂y*  ∂y*  ∂y h2 ∂y*2
Thus, the original differential equation becomes
ρU ω
μU ∂ 2 u*
=
∂t*
h2 ∂y*2
∂u*
or
 ρω h2  ∂u* ∂ 2 u*

 * = *2
∂y
 μ  ∂t
Problem 7.84
The deflection of the cantilever beam of the figure below is governed by the differential
equation.
d2y
= P (x − )
dx 2
where E is the modulus of elasticity and I is the moment of inertia of the beam cross
section. The boundary conditions are y = 0 at x = 0 and dy / dx = 0 at x = 0. (a) Rewrite the
equation and boundary conditions in dimensionless form using the beam length, , as the
EI
reference length. (b) Based on the results of part (a), what are the similarity requirements
and the prediction equation for a model to predict deflections?
y
P
x
ℓ
Solution 7.84
y
(a) Let y* =
and
=
so that:
( )
*
dy d y dx*
dy*  1  dy*
=
=
 =
dx
dx* dx
dx*   dx*
and
d2y
dx
=
2
d  dy*  dx* 1 d 2 y*
=


dx*  dx*  dx
dx*2
Thus, the original differential becomes
2 *
(
 EI  d y
*
  dx*2 = P x −
)
or
P 2  *
=
 x −1
dx*2  EI 
d 2 y*
(
)
and the boundary conditions are
dy*
y* = 0 at x* = 0 and
= 0 at x* = 0
*
dx
(b) The similarity requirement are
*
xm
= x * or
xm x
P 2
P 2
= and m m =
Em I m
EI
m 
The prediction equation is
*
y* = ym
or
y ym
=
 m
Problem 7.85
A liquid is contained in a pipe that is closed at one end as shown in the figure below.
Initially the liquid is at rest, but if the end is suddenly opened the liquid starts to move.
Assume the pressure p1 remains constant. The differential equation that describes the
resulting motion of the liquid is
 ∂2vz 1 ∂ vz 
∂vz p1
ρ
= +μ 2 +

 ∂r
r ∂r 
∂t

where vz is the velocity at any radial location, r, and t is time. Rewrite this equation in
dimensionless form using the liquid density, ρ; the viscosity, μ; and the pipe radius, R, as
reference parameters.
p1
End initially
closed
vz
r
z
R
ℓ
Solution 7.85
v*
r * t
, t = , and vz* = z where is some combinations of the parameters ρ, μ, and R
R
τ
V
having the dimensions of time, and V is some combination of the same parameters having
the dimensions of a velocity. Let
Let r* =
( −4 2 )
FL T ( L )
ρ R2
τ=

T
μ
FL−2T
2
and
V=
μ
FL−2T

 LT −1
4
2
−
ρR
FL T ( L )
(
)
With these dimensionless variables:
( *)
∂vz ∂ Vvz ∂t*
∂vz*  1   μ   μ  ∂vz*  μ  1 ∂vz*
=
=V *   = 
= 


∂t
∂t* ∂t
∂t  τ   ρ R   ρ R 2  ∂t*  ρ  R3 ∂t*
(
∂vz ∂ Vvz
=
∂r
∂r*
*
2
) ∂r* = V ∂vz*  1  = μ  1  ∂v*z = μ ∂vz*
∂r
 
∂r*  R 
ρ R  R  ∂r*
ρ R 2 ∂r*
∂  ∂v*z  ∂r*
μ ∂ 2vz*
μ ∂ 2vz*  1 
=
=


 
∂r
ρ R 2 ∂r*  ∂r*  ∂r ρ R 2 ∂r*2  R  ρ R3 ∂r*2
The original differential equation can be expressed as
∂ 2ν z
=
2
μ
  μ 2 1  ∂v* p   μ    ∂ 2ν * 1 ∂v* 
z +
z
 ρ   3  *z = 1 +  μ 

3  
*2
*
*



r ∂r 
  ρ  R  ∂t
  ρ R    ∂r
or
∂v*z
p1ρ R3
∂t
μ2
=
*
+
∂ 2vz*
1 ∂vz*
+
∂r*2 r* ∂r*
Problem 7.86
A student drops two spherical balls of different diameters and different densities. She has a
stroboscopic photograph showing the positions of each ball as a function of time. However,
she wants to express the velocity of each as a function of time in dimensionless form.
Develop the dimensionless groups. The equation of motion for each ball is:
C
dV
mg − D ρ AV 2 = m
,
2
dt
where m is ball mass, g is acceleration of gravity, CD is a dimensionless and constant drag
2
coefficient, ρ is air mass density, A is ball cross-sectional area (= π D 4) with D ball
diameter, V is ball velocity, and t is time.
Solution 7.86
CD
dV
ρ AV 2 = m
dt
2
First put the equation in nondimensional form by dividing by mg
mg −
1−
Note that
A=
π
CD ρ AV 2 1 dV
=
g dt
2 mg
D 2 and m = ρs V = ρs
π
D3
4
6
where ρS is the density of the sphere material. Substituting and simplifying on the left-hand
side of the equation gives
3
ρ V 2 1 dV
1 − CD
=
ρs gD g dt
4
On the right-hand side of the equation, multiply and divide by gD and
3
ρ V
1
g
1 − CD
=
gD
4
D
ρs gD g
2
d
V
gD
g
d
t
D
d
V
gD
d
gt 2
D
=
g
to get
D
gt 2
V
as a dimensionless time, the
is the Froude Number, Fr. Defining t ≡
D
gD
differential equation can be written as
3
ρ
d (Fr)
= 1 − CD (Fr)2
ρs
4
dt
The term
There are four dimensionless parameters: CD ;
ρ
; Fr; t defined as above.
ρs
Problem 7.87
The basic equation that describes the motion of the fluid above a large oscillating flat
plate is
∂u
∂2u
=v 2 ,
∂t
∂y
where u is the fluid velocity component parallel to the plate, t is time, y is the spatial
coordinate perpendicular to the plate, and v is the fluid kinematic viscosity. The plate
oscillating velocity is given by U = U0 sin(ωt). Find appropriate dimensionless parameters
and the dimensionless differential equation.
Solution 7.87
∂u
∂2u
=v 2
∂t
∂y
Fluid
Plate
u = U0 sin ω t
We will nondimensionalize the equation by introducing reference value for the variables.
1
The reference value for u is obviously U0 and the reference value for t is obviously . There
ω
is no reference length per se; however,
 L  
L
 

U
U
 0
T 
0
T
 ω  =   1   = [L] . ω  1 = L
  

  T  
T
Introduce
u′ =
u
yω
, t′ = ωt, y′ =
.
U0
U0
Then
U 0ω
∂u′ vU 0ω 2 ∂ 2 u′
and
=
∂t′
U 2 ∂y′2
The appropriate dimensionless parameters are
u
yω
vω
, ωt,
, and 2 .
U0
U0
U0
∂u′ vω ∂ 2 u′
.
=
∂t′ U 02 ∂y′2
Problem 7.88
The dimensionless parameters for a ball released and falling from rest in a fluid are
gt2
ρ
Vt
CD ,
,
, and
,
ρb
D
D
where CD is a drag coefficient (assumed to be constant at 0.4), g is the acceleration of
gravity, D is the ball diameter, t is the time after it is released, is the density of the fluid in
which it is dropped, and is the density of the ball. Ball 1, an aluminum ball b1 =
2710kg/m3 having a diameter D1 = 1.0 cm, is dropped in water ( = 1000 kg/m3). The ball
velocity V1 is 0.733 m/s at t1 = 0.10 s. Find the corresponding velocity V2 and time t2 for
ball 2 having D2 = 2.0 cm and ρ b2 = ρ b1 . Next, use the computed value of V2 and the
equation of motion,
C
dV
,
ρbVg − D ρ AV 2 = ρbV
2
dt
where V is the volume of the ball and A is its cross-sectional area, to verify the value of t2.
Should the two values of t2 agree?
Solution 7.88
The dimensionless group
gt 2
gives
D
 gt22   gt12 

 = 
 ,
D
D
2
1

 

D 
t2 = t2  2  ,
 D1 
 2.0 cm 
t2 = ( 0.10 s ) 
 , or
 1.0 cm 
Vt
The dimensionless group
gives
D
 V2t2   V1t1 
 t  D 
V2 = V1  1   2  ,

=
,
 D2   D1 
 t2   D1 
m   0.10 s   2.0 cm 

V2 =  0.733  


s   0.141s   1.0 cm 

or
V2 = 1.04
m
s
Now write Eq. (1) as
g−
CD ρ AV 2 dV
=
,
2 ρb V
dt
Noting
A π D2 6
3
=
=
,
4 π D3 2D
V
t2 = 0.141s
rearranging, and integrating gives
V
t
dV
=
0 1 − α 2V 2 0 gdt
3ρ
where α 2 =
. Expanding the left-hand side by partial fractions and carrying out the
4 ρ b gD
integration gives
1 V dV
1 V dV
+ 
= gt,

2 0 1 − αV 2 0 1 + αV
1
1
−
ln (1 − αV ) +
ln (1 + αV ) = gt,
2α
2α
or
 1 + αV 
ln 
.
2α g  1 − αV 
For the D2 = 2.0 cm ball
t=
1
kg 

3 ( 0.4 )  1000 3 
3CD ρ
s
m 

α=
=
= 0.751
kg  
m  2
4 ρb gD
m


4  2710 3   9.81 2  
m

m 
s   100 
1
1
=
= 0.0679s,
s 
m
2α g

2  0.751  9.81 2 
m 

s 
and
 
s  
 1 +  0.751 m V 
 .
t = 0.0679s ln  
 1 −  0.751 s V 

 
m  
 
For V = V2 = 1.04
m
,
s
 1 + (0.751×1.04) 
t2 = 0.0679 s ln 

 1 − (0.751× 1.04) 
or
t2 = 0.142 s,
close to 0.141s.
These two values of t2 should agree since both models (dimensional analysis and differential
equation) model the same physical system.
Problem 8.2
Rainwater runoff from a parking lot flows through a 3-ft- diameter pipe, completely filling
it. Whether flow in a pipe is laminar or turbulent depends on the value of the Reynolds
number. Would you expect the flow to be laminar or turbulent? Support your answer with
appropriate calculations
Solution 8.2
Re =
ρVD VD
If Re > 4000 the flow is turbulent. The corresponding velocity is
=
μ
ν

ft 2 
(4000) 1.21×10−5


s 
ReV
ft

=
= 0.161
V=
3ft
s
D
Most likely the velocity will be greater than this, that is, turbulent flow.
Problem 8.3
Blue and yellow streams of paint at 60 F (each with a density of 1.6
slugs
and a viscosity
ft 3
ft
1000 times greater than water) enter a pipe with an average velocity of 4 as shown in the
s
figure below. Would you expect the paint to exit the pipe as green paint or separate streams
of blue and yellow paint? Explain. Repeat the problem if the paint were “thinned” so that it
is only 10 times more viscous than water. Assume the density remains the same
Yellow
Green?
2 in.
Splitter
Blue
25 ft
Solution 8.3
If the flow is laminar, the paint would exit as separate blue and yellow streams.
slugs  ft   2 
4
ft



ρVD
ρVD
ft 3  s   12  = 45.6 < 2100
=
=
Re =
1000 μ H O 1000 2.34 × 10 −5 lb ⋅ s
μ
1.6
(
2
) ft 2
Thus, laminar flow so blue and yellow streams.
If using μ = 10 μH2O obtain
Re = 4560 > 4000 , so have turbulent flow with natural mixing and green paint.
Note: Check to determine if the 25 ft length is greater than the entrance length,
For laminar flow,
e
D
For turbulent flow,
= 0.06 Re , or
e
D
1
6
= 4.4 Re , or
2 
ft  = 0.456 ft < 25 ft
 12 
e = 0.06(45.6) 
e
1
 2 
6
= 4.4(4560)
ft = 2.99 ft < 25 ft

 12


e .
Problem 8.4
Air at 200 F flows at standard atmospheric pressure in a pipe at a rate of 0.08
lb
.
s
Determine the minimum diameter allowed if the flow is to be laminar.
Solution 8.4
Maximum Re =
ρVD
for laminar flow is Re = 2100.
μ
or with
V=
Q
4Q
,Re =
=
A π D2
 4Q 
D
 π D 2  = 4 ρQ = 2100
μ
πμ D
ρ
Hence,
Q=
2100πμ D
4ρ
Given γ Q = 0.08
(1)
p
lb
, where γ = g ρ and ρ =
RT
s
Thus,
lb 

14.7 × 144 2 
slugs
ft 

= 0.00187 3
ρ=
ft ⋅ lb
ft
1716
460 + 200 ) ° R
° (
slug ⋅ R
so that
0.08
lb
s
ft3
Q=
= 1.33
ft  
slugs 
s

 32.2 2   0.00187 3 
s 
ft 

lb ⋅ s
Hence, with μ = 4.49 × 10 −7 2 (see Table B.4 Physical Properties of Air at Standard
ft
Atmospheric Pressure [BG/EE Units]),
Eq. (1) gives
slugs  
ft 2 

4  0.00187 3  1.33

s 
ft  

4 ρQ
=
= 3.36 ft
D=
2100πμ

−7 lb ⋅ s 
2100π  4.49 ×10

ft 2 

Problem 8.5
ft3
To cool a given room, it is necessary to supply 4
of air through an 8-in.- diameter pipe.
s
Approximately how long is the entrance length in this pipe?
Solution 8.5
ft3
4
2
Q
ft
−4 ft
s
11.5
V= =
=
ν
=
×
1.57
10
Thus,
with
s
A π  8 2
s
 ft 
4  12 
(see Table B.4 Physical Properties of Some Common Gases at Standard
Atmospheric Pressure [BG Units])
ft  8 
 ft 
VD
s  12 
=
= 48800 > 4000 so the flow is turbulent.
Re =
2
ν
−4 ft
1.57 ×10
s
11.5
Hence,
e
D
1
6
= 4.4 Re , or
e
1
8
6
= 4.4(48800)
= 17.7 ft
 
 12 
Problem 8.6
The flow of water in a 3-mm- diameter pipe is to remain laminar. Plot a graph of the
maximum flowrate allowed as a function of temperature for 0 < T < 100 C .
Solution 8.6
For laminar flow, Re=
VD
ν
≤ 2100, where V =
Q
4Q
=
A π D2
Thus, the maximum Q is given by
 4Q 
D

2
2100πν D
4Q
π
D


Re =
=
= 2100, or Q =
4
ν
πν D
or
Q=
m2
m3
2100πν D(0.003m)
and Q ∼
ν = 4.95ν , where ν =
s
s
4
With values of ν from Table B.2 Physical Properties of Water (SI Units), we obtain
T (°C)
 m2 
V

 s 
 m3 
Q

 s 
0
20
40
60
80
100
1.79 × 10−6
1.00 × 10−6
6.58 × 10−7
4.75 × 10−7
3.65 × 10−7
2.90 × 10−7
8.86 × 10−6
4.95 × 10−6
3.26 × 10−6
2.35 × 10−6
1.81 × 10−6
1.44 × 10−6
10
106Q,
m3
s
5
0
0
50
T, °C
100
Problem 8.7
The pressure distribution measured along a straight, horizontal portion of a 50-mm- diameter
pipe attached to a tank is shown in the table below. Approximately how long is the entrance
length? In the fully developed portion of the flow, what is the value of the wall shear stress?
x (m) (±0.01 m)
0 (tank exit)
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0 (pipe exit)
p (mm H2O) (±5 mm)
520
427
351
288
236
188
145
109
73
36
0
Solution 8.7
The entrance length extends to the fully developed portion in which
Approximate
∂p δ p
≈
to obtain the following:
∂x δ x
∂p
= constant.
∂x
From x =
to x = () m
∂p, mm H2O
∂x
∂p mmH 2 O
,
m
∂x
0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
4.5
5.0
−93
−76
−63
−52
−48
−43
−36
−36
−37
−36
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
0.5
−186
−152
−126
−104
−96
−86
−72
−72
−74
−72
Within the error on δ p, the pressure gradient is constant for x ≥ 3m . Thus
e ≈ 3m .
N 
N
mm H 2O

Since 1mm H2O × γ H2O = (1× 10−3 m)  9800 3  = 9.80 2 , then
m
m 
m

N 

9.80 2 
mm H 2O 
Δp
m  = 706 N
= 72

m
m3
 mm H 2O 




For x > 3m ,
Δp
= 72
4τ w
if follows that
D
D Δp 0.050 m 
N 
N
=
τw =
 706 3  = 8.83 2
4
4
m 
m

Since Δp =
Problem 8.8
Nanoscale flows The term nanoscale generally refers to objects with characteristic lengths
from atomic dimensions up to a few hundred nanometers (nm). (Recall that 1 nm = 10−9
m.) Nanoscale fluid mechanics research has recently uncovered many surprising and useful
phenomena. No doubt many more remain to be discovered. For example, in the future
researchers envision using nanoscale tubes to push tiny amounts of water-soluble drugs to
exactly where they are needed in the human body. Because of the tiny diameters involved,
the Reynolds numbers for such flows are extremely small and the flow is definitely laminar.
In addition, some standard properties of everyday flows (e. g., the fact that a fluid sticks to
a solid boundary) may not be valid for nanoscale flows. In addition, ultratiny mechanical
pumps and valves are difficult to manufacture and may become clogged by tiny particles
such as biological molecules. As a possible solution to such problems, researchers have
investigated the possibility of using a system that does not rely on mechanical parts. It
involves using light-sensitive molecules attached to the surface of the tubes. By shining light
onto the molecules, the light-responsive molecules attract water and cause motion of water
through the tube. (See Problem 8.8.)
(a) Water flows in a tube that has a diameter of D = 0.1m . Determine the Reynolds
number if the average velocity is 10 diameters per second. (b) Repeat the calculations if the
tube is a nanoscale tube with a diameter of D = 100 nm .
Solution 8.8
2
10(0.1m)
m
−6 m
(a) Re =
, where D = 0,1m , V =
= 1 , and ν = 1.12 ×10
ν
s
s
s
VD
Thus,
 m
1  (0.1m)
s 
Re = 
= 89300
2
−6 m
1.12 × 10
s
(b) Re =
10(10−7 m)
 1m 
−6 m
−7
=
=
V
10
D
=
=
100
nm
10
m
, where
,
,


ν
s
s
 109 nm 
VD
and ν = 1.12 ×10
2
−6 m
s
Thus,
 −6 m  −7
10
 (10 m)
s 

Re =
= 8.93 ×10−8
2
m
1.12 ×10 −6
s
Problem 8.9
For fully developed laminar pipe flow in a circular pipe, the velocity profile is given by

m
r2 
u( r ) = 2 1 − 2  in
, where R is the inner radius of the pipe. Assuming that the pipe
 R 
s


diameter is 4 cm , find the maximum and average velocities in the pipe as well as the
volume flowrate.
Solution 8.9

r2 
u( r ) = 2  1 − 2 
 R 


  2 r 2 
 ΔpD 2    2r 2 
Based on the equation u( r ) = 
= Vc 1 −    ,
1−
 16 μ    D  



  D  
Maximum velocity, Vc = 2
m
s
We could also use the fact that the maximum velocity occurs at the centerline of the pipe,
r=0
0 
m

u(0) = Vc = 2  1 − 2  = 2
s
 R 
Average velocity, V
V=
Vc 2
m
= =1
2 2
s
Volume flowrate, Q
π
2
Q = VA = (1) (0.04) = 1.26 ×10
4
3
−3 m
s
Problem 8.10
A viscous fluid flows in a 0.10-m- diameter pipe such that its velocity measured 0.012 m
m
away from the pipe wall is 0.8 . If the flow is laminar, determine the centerline velocity
s
and the flowrate.
Solution 8.10
For laminar flow in a pipe,
  2 r 2 
0.1m
m
− 0.12 m = 0.038m
at r =
u( r ) = Vc 1 −    , where D = 0.1 m and u = 0.8
2
s
  D  
Thus,
0.8
 2(0.038m)2 
m
m
= Vc 1 −
 or Vc = 1.89
s
0.10 m 
s

so that
Q=
π
4
D2V =
π
4
D2 ( 0.5Vc ) =
π
3
m
m

(0.1m)2 (0.5) 1.89  = 7.42 ×10−3
4
s 
s

Problem 8.11
The wall shear stress in a fully developed flow portion of a 12-in. -diameter pipe carrying
∂p
lb
water is 1.85 2 . Determine the pressure gradient,
, where x is in the flow direction, if
∂x
ft
the pipe is (a) horizontal, (b) vertical with flow up, or (c) vertical with flow down.
Solution 8.11
In general,
Δp − γ sin θ
=
Thus, with τ ≈ τ w at r =
2τ
r
D
∂p
Δp
=−
and
, this becomes
2
∂x
4τ
∂p
= − w − γ sin θ
∂x
D
a) For a horizontal pipe θ = 0
lb 

4 1.85 2 
4τ
∂p
ft  = −7.40 lb
=− w =− 
∂x
D
1ft
ft3
b) For a vertical flow up θ = 90
lb 

4 1.85 2 
4τ
∂p
ft  − 62.4 lb = −69.8 lb
= − w −γ = − 
1ft
∂x
D
ft3
ft3
and
c) For a vertical down θ = −90
lb 

4 1.85 2 
4τ
∂p
ft  + 62.4 lb = 55.0 lb
= − w +γ = − 
1ft
∂x
D
ft3
ft3
Problem 8.12
The pressure drop needed to force water through a horizontal 1-in. -diameter pipe is
0.60 psi for every 12-ft length of pipe. Determine the shear stress on the pipe wall.
Determine the shear stress at distances 0.3 and 0.5 in. away from the pipe wall.
Solution 8.12
For a horizontal pipe,
Δp
=
2τ
r Δp
or τ =
r
2
lb 

 0.6 ×144 2 
ft  = 3.6r lb , where r ∼ ft
Thus, τ = r 
2(12 ft)
ft 2
Hence,
lb
 0.5 
 = 0.15 2
 12 
ft
τ w = 3.6 
and with r = ( 0.5 − 0.3) in. = 0.2in. ,
lb
 0.2 
 = 0.06 2
 12 
ft
τ = 3.6 
Finally, with r = (0.5 − 0.5)in. = 0 in. τ = 0
Problem 8.13
The pressure drop needed to force water through a 1-in. -diameter pipe is 0.60 psi for every
12-ft length of pipe. The pipe is on a 20 hill. Is the flow, up or down the hill? Explain.
Determine the shear stress on the pipe wall. Determine the shear stress at distances 0.3 and
0.5 in. away from the pipe wall.
Solution 8.13
For a pipe on a hill
Δp
=
2τ
+ γ sin θ , where θ = ±20
r
Assume the flow is uphill: θ = +20


lb 
0.6 × 144 2 



r  Δp

1 0.5  
ft  − 62.4 lb sin 20 
Thus, τ =  − γ sin θ  or τ w = 
ft  
2
2  12  
12ft


ft 3


or
τ w = −0.295
lb
ft 2
Since we must have τ w > 0, the flow must not be uphill.
Assume the flow is downhill: θ = −20


lb 
0.6 × 144 2 



r  Δp

r
ft  + 62.4 lb sin 20 
Thus, τ =  − γ sin θ  or τ w =  
2
2
12ft

ft 3



= 14.3r
lb
ft 2
, where r ∼ ft . The flow is downhill.
Hence, with r =
D
2
lb
 0.5 
 = 0.596 2
 12 
ft
τ w = 14.3 
with r = ( 0.5 − 0.3) in. = 0.2in. ,
lb
 0.2 
 = 0.238 2
 12 
ft
τ w = 14.3 
With r = (0.5 − 0.5)in. = 0, τ = 0
Problem 8.14
Water flows in a constant-diameter pipe with the following conditions measured: At section
(a) pa = 32.4 psi and za = 56.8 ft ; at section (b) pb = 29.7 psi and zb = 68.2 ft . Is the flow is
from (a) to (b) or from (b) to (a)? Explain.
Solution 8.14
?
(b)
(a)
Va 2
pb Vb 2
Assume the flow is uphill. Thus,
+
+ za =
+
+ zb + hL
γ
2g
γ
2g
pa
or with Va = Vb ,
The head loss is given by

in.2 
(32.4 psi − 29.7 psi ) 144 2 
ft 
pa
pb

+ za −
− zb =
+ 56.8ft − 68.2 ft
hL =
lb
γ
γ
62.4 3
ft
or
hL = −5.17ft < 0, which is impossible. Thus, the flow is downhill, from (b) to (a).
Problem 8.15
For laminar flow in a round pipe of diameter D , at what distance from the centerline is the
actual velocity equal to the average velocity?
Solution 8.15
D
u(r)
r
Vc
For laminar flow,
  2r 2 
u = Vc 1 −   
  D  
  2r 2 
Vc
D
Thus, if u =
= 0.354D
= Vc 1 −    , r =
2
2 2
  D  
Problem 8.16
Glycerin at 20 C flows upward in a vertical 75-mm- diameter pipe with a centerline
m
velocity of 1 . Determine the head loss and pressure drop in a 10-m length of the pipe.
s
• Z 2 = Z1 t
(2)
Solution 8.16
ρ = 1260
μ = 1.50
kg
m3
N ⋅s
m2
ℓ = 10 m
For laminar flow in a pipe,
1
2
V = average velocity= Vmax =
1 m
m
1  = 0.5
2 s 
s
• Z1
(1)
Thus,
kg 
m

1260 3  0.5  ( 0.075 m )

ρVD 
s 
m 
=
= 31.5 < 2100
Re =
N ⋅s
μ
1.50 2
m
The flow is laminar so that
V=
( Δp − γ sinθ ) D2 ,
32 μ
where θ = 90
Thus,
N ⋅s 
m


32 1.50 2  (10 m )  0.5 
32 μ V
m 
kg 
s  
m 

Δp =
+γ = 
+  9.81 2  1260 3  (10 m )
2
2
D
s 
m 

( 0.075 m )
= 1.66 × 105
N
m2
, or Δp =166 kPa
In addition,
V12 p2
V2 2
+ z1 +
=
+ z2 +
+ hL , or with V1 = V2 , z2 − z1 = , and
γ
2g
γ
2g
p1
p1 = p2 + Δp this gives
N
m2
− =
− 10 m = 3.43 m
hL =
m 
kg 
γ

 9.81 2  1260 3 
s 
m 

Δp
D = 0.0
1.66 ×105
Problem 8.17
Water at 60 °F flows at a rate of 4.0 gal /min through a 6-in. I.D. plastic pipe. The pipe is
500 ft long and rises a vertical height of 40 ft over the 500 ft . Find the pressure drop.
Solution 8.17
The Reynolds number is
gal   ft3   min 

4  4.0



min   7.48gal   60 s 

4Q
Re =
=
= 1876 ,
π Dν
6 
−5 2
π  ft  1.21×10 ft /s
 12 
(
)
so we have laminar flow. The energy loss for fully developed flow is
2

gal 

−5 ft 
128 1.21×10
 ( 500 ft )  4.0


s 
min 

128γ Q

=
ghL =
4
π D4
s 
gal  slug ⋅ ft 
6  
π  ft   60
 7.48 3 

 12   min  
ft  lb ⋅ s2 
ghL = 0.0352 ft ⋅ lb/slug.
The pressure loss is found using the density of water.
slugs  
ft ⋅ lb 

Δ pL = ρ ghL = 1.938 3   0.0352

slug 
ft  

Δ pL = 0.068 lb/ft 2 .
The pressure drop is found by writing the mechanical energy equation from the inlet (1) to
the outlet (2).
p1 V12
p2 V22
+
+ gz1 =
+
+ gz2 + ghL
ρ
2
ρ
2
or
p1 − p2 = ρ g ( z2 − z1 ) + ρ ghL = ρ g ( z2 − z1 ) + pL
 lb ⋅ s 2 
slug  
ft 
lb

=  1.938 3   32.2 2  ( 40 ft ) 
+ 0.069 2

 slug ⋅ ft 
ft  
s 
ft



p1 − p2 = 2500
lb
ft 2
= 17.3psi.
Problem 8.18
At time t = 0 , the level of water in tank A shown in the figure below is 2 ft above that in
tank B . Plot the elevation of the water in tank A as a function of time until the free
surfaces in both tanks are at the same elevation. Assume quasi-steady conditions—that is,
the steady pipe flow equations are assumed valid at any time, even though the flowrate does
change (slowly) in time. Neglect minor losses. Note: Verify and use the fact that the flow is
laminar.
2 ft at t = 0
3 ft
3 ft
25 ft
B
A
0.1-in. diameter, galvanized iron
Solution 8.18
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
, where p1 = p2 = 0 and V1 = V2 ≈ 0
2g
D 2g
γ 2g
γ
(1)
At t = 0, z1 = 0 and z1 = h0 = 2 ft
Δ1
Z
(2)
•
Δ2
h0 = 2 ft
•
(1)
V
Because the tanks are the same diameter
Δ1 = Δ 2 and with z2 = Δ 2 , z1 = h0 − Δ 2
we obtain z1 = h0 − z2 Thus, Eq. (1) becomes
V2
V2
z1 = z2 + f
or 2 z1 − h0 = f
D 2g
D 2g
(2)
π
π
 dz 
Also, A1  − 1  = Q = D2V , where A1 = DT 2 with DT = 3ft= tank diameter
4
4
 dt 
2
 D  dz1
Thus, V = −  T 
 D  dt
The maximum Re =
(3)
ρVD
occurs when the head, z1 − z2 , is greatest.
μ
Vmax 2
From Eq. (2) (with z1 = h0 ), h0 = f
D 2g
Assume laminar flow so that f =
64
64 μ
or f =
Re
ρVD
(4)
Thus, from Eq. (4)
h0 =
Vmax 2 32 μ Vmax
64 μ
, or
=
ρVmax D D 2 g
γ D2
2
lb  0.1 

62.4 3  
ft  ( 2 ft )
2

γ D h0
ft
12 

ft


Vmax =
=
= 0.462
lb ⋅ s 
32 μ
s

32  2.34 ×10−5 2  ( 25ft )
ft 

or
slugs 
ft  0.1 

1.94 3   0.462 s  12 ft 

 = 319 < 2100
ft  
Remax = 
The flow remains laminar.
⋅
lb
s
−5
2.34 ×10
ft 2
Thus, Eqs. (2) and (4) give
2 z1 − h0 =
64 μ V 2 32 μ V
=
, or using Eq. (3)
ρVD D 2 g
γ D2
2
 D  32 μ dz1
2 z1 − h0 = −  T 
 D  γ D2 dt
(5)
h
dF dz1
=
and Eq. (5) becomes
Let F = z1 − 0 so that
2
dt
dt
2
 D  32 μ dF
2F = −  T 
 D  γ D2 dt
or α
dF
16 μ  DT 
+ F = 0 , where α =


dt
γ D2  D 
2
Thus,
α
dF
= −  dt or α ln F = −t + C , where C = constant
F
Hence,
t
− 
F = Ce  α 
C=
t
− 
h
That is, z1 − 0 = Ce  α  with the initial condition z1 = h0 when t = 0, or
2
h0
2
t
h
h − 
Thus, z1 − 0 = 0 e  α 
2
2
or
t

−  
h
h0 
1 + e  α   Note: As t → ∞, z1 → 0
z1 =

2
2 


For the conditions given, h0 = 2 ft and
2
lb ⋅ s 



16  2.34 ×10 −5 2  ( 25 ft ) 
3ft 
ft 

 = 2.80 ×105 s
α= 
2
  0.1 ft  
lb   0.1 

  12  
 62.4 2   12 ft 



ft  

Hence,
z1
 t

−

2.8×105 

= 1+ e
, where z
1 ∼ ft and t ∼ s
This result is plotted below. (Note: lim z1 = 1 ft )
t →∞
2.5
2
z (ft)
1.5
1
0.5
0
0.E+00
2.E+05
4.E+05
6.E+05
t (s)
8.E+05
1.E+06
Problem 8.19
A fluid flows through a horizontal 01-in.- diameter pipe. When the Reynolds number is
1500 , the head loss over a 20-ft length of the pipe is 6.4 ft . Determine the fluid velocity.
Solution 8.19
V2
hL = f
, where since Re = 1500 < 2100 the flow is laminar.
D 2g
Thus, f =
64
64
=
= 0.0427 so that
Re 1500
6.4 ft = 0.0427
or
V = 2.01
ft
s
V2
20 ft
ft 
 0.1  
ft  2  32.2 2 

 12  
s 
Problem 8.20
Asphalt at 120 F, considered to be a Newtonian fluid with a viscosity 80000 times that of
water and a specific gravity of 1.09 , flows through a pipe of diameter 2.0 in. If the pressure
psi
gradient is 1.6
determine the flowrate assuming the pipe is (a) horizontal; (b) vertical
ft
with flow up.
Solution 8.20
π ( Δp − γ sin θ ) D 4
If the flow is laminar, then Q =
128μ
lb 
lb

where γ = SGγ H2O = 1.09  62.4 3  = 68.0 3
ft 
ft

and


μ = 80000μH O = 8 ×104 1.164 ×10−5
2
lb ⋅ s 
lb ⋅ s
= 0.931 2
2 
ft 
ft
a) For horizontal flow, θ = 0
Thus, from Eq. (1)
4
lb  2 

π 1.6 ×144 2   ft 
3
12 
ft



−3 ft
Q=
= 4.69 ×10
lb ⋅ s 
s

128  0.931 2  (1ft )
ft 

b) For vertical flow up, θ = 90
Thus, from Eq. (1)
4


lb
lb
 2 
68
1ft
−
(
)
  12 ft 
3
2
3
ft
ft
 = 3.30 ×10−3 ft

Q=
lb ⋅ s 
s

128  0.931 2  (1ft )
ft 

Note: We must check to see if our assumption of laminar flow is correct.
π 1.6 ×144
ft3
Q
s = 0.215 ft , it follows that
Since V = =
2
s
A
π2
 
4  12 
slug 

2 
1.09 1.94 3  ( 0.215 )  ft 
ρVD
ft 
 12  = 0.0814 < 2100

Re =
=
lb ⋅ s
μ
0.931 2
ft
The flow is laminar.
4.69 ×10−3
(1)
Problem 8.21
Oil of SG = 0.87 and a kinematic viscosity ν = 2.2 ×10
−4
shown in the figure below at a rate of 4 ×10
20 mm
4m
h
SG = 1.3
Q
Solution 8.21
SG = 0.87
(1)•
γ
20 mm
h1
h2
h
(2)•
Q
γm
SG = 1.3
m3
Q
s = 1.27 m so that
V= =
A π 0.02m 2
s
(
)
4
m

1.27  ( 0.02m )

ρVD VD 
s 
=
=
= 115 < 2100
Re =
2
μ
ν
−4 m
2.2 ×10
s
4 ×10−4
m2
flows through the vertical pipe
s
m3
. Determine the manometer reading, h .
s
SG = 0.87
ℓ=4m
−4
The flow is laminar with
Q=
π ( Δp + γ ) D 4
128μ Q
, or Δp = p1 − p2 =
−γ
128μ
π D4
(1)
kN 
kN

Hence, with γ = SGγ H2O = 0.87  9.81 3  = 8.53 3 and
m 
m

2

kg 
N ⋅s

−4 m 
μ = νρ = ν SG ρH2O =  2.2 ×10
 ( 0.87 ) 1000 3  = 0.191 2
s 
m 
m


Eq. (1) gives

N ⋅s 
m3 

128  0.191 2  ( 4 m )  4 ×10−4


s  

m 
kN 
 3 N 

Δp =
−
8.53
4
m
(
)



10
4
kN 


m3 
π ( 0.020m )
or
Δp = 4.37 ×10 4
N
m
2
= 43.7
kN
(2)
m2
From manometer considerations,
kN 
kN

p1 + γ h1 − γ m h + γ h2 = p2 , where γ m = SGmγ H2O = 1.3  9.81 3  = 12.74 3
m 
m

and h1 = h − h2 + , or h2 + h1 = h +
Thus,
p1 − p2 = Δp = −γ ( h2 + h1 ) + γ m h = ( γ m − γ ) h − γ
(3)
Combine Eqs. (2) and (3) to give
43.7
kN
m
= (12.74 − 8.53)
2
kN 

h
8.53
−

 ( 4 m ) or
m3
m3 

kN
h = 18.5 m
Problem 8.22
A liquid with SG = 0.96 , μ = 9.2 × 10 −4
N ⋅s
, and vapor pressure pv = 1.2 × 10 4
N
(abs) is
m
m2
drawn into the syringe as is indicated in the figure below. What is the maximum flowrate if
cavitation is not to occur in the syringe?
2
10-mm-diameter
0.12 m
0.25-mm-diameter
0.10-m-long needle
patm = 101 kPa (abs)
Solution 8.22
2
p1 V12
p V2

V
+
+ z1 = 2 + 2 + z2 +  f +  K L 
, where p1 = 101 kPa, z1 = 0
γ 2g
γ
2g
 D
 2g
V1 = 0, z 2 = 0.12 m. The maximum flowrate will occur when p2 is the minimum allowed:
p2 = pv = 1.2 × 10 4
Thus,
p1
γ
=
pv
γ
+
N
m2
.
2
V2 2

V
+ z2 +  f + K Lentrance + K Lexit 
,
2g
 D
 2g
2
(1)
2
 D 
 0.25 mm 
VA
where V2 =
=V 
 =V 
 = 0.000625V Thus, V2 ≈ 0
A2
 10 mm 
 D2 
and Eq. (1) becomes
(101×103 −1.2 ×104 ) mN2
N 

0.96  9.80 ×103 3 

m 
or
122 = ( 267 f + 1)V 2
 

0.1m
V2

0.5
1
= 0.12 m  f 
+
+

−3 
  0.25 ×10 m 
 2  9.81 m 


s2 

(2)
Assume (because of the small diameter) that the flow is laminar.
Thus, f =
64 64 μ
=
Re ρVD
or
N ⋅s 

64  9.2 ×10 −4 2 
0.246
m 

F=
=
kg 
V

0.96  999 3 V 0.25 × 10−3 m
m 

Hence, from Eq. (2),
(
)
0.246  2

122 =  267 +
+ 1V or 122V = ( 65.7 +V )V 2
V


Thus,
V 2 + 65.7V − 122 = 0, which has solutions
V=
−65.7 ± 65.72 + 4 (122 )
2
= 1.81
m
m
, or −65.7
(neglect the V < 0 root)
s
s
Hence,
Q = AV =
π
m
m

(0.25 × 10 −3 m)2  1.81  = 8.88 × 10 −8
4
s 
s

Check if laminar flow:
(
)
kg 
m

0.96  999 3 1.81  0.25 ×10−3 m
ρVD
s 
m 

Re =
=
= 472 < 2100 (laminar)
μ
−4 N ⋅ s
9.2 ×10
m2
Problem 8.23
A 3-in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in. )
carries 210 °F SAE 40 crankcase oil at the rate of 6.0 gal/min . The oil specific gravity is
−7
2
0.89, and the absolute viscosity is 6.6 ×10 lb ⋅ s / ft . For the same pressure drop, what must
the new pipe size be to carry the oil at 10.7 gal/min ?
Solution 8.23
Apply the mechanical energy equation from 1 to 2.
p1 V12
p2 V2 2
+
+ gz1 − ws =
+
+ gz2 + ghL
ρ
2
ρ
2
Δp
•1
Now ws = 0 and, assuming constant fluid density, the
2•
ℓ
continuity equation gives V1 = V2 . This gives
p1 − p2 = ρ g ( z2 − z1 ) + ρ ghL .
Since both the 3-in. pipe and the larger pipe have the same pressure drop ( p1 − p2 ) ,
we write
( p1 − p2 )3in. = ( p1 − p2 )d
where the d subscript represents the larger pipe diameter. Then
 f V2 
 f V2 
= ρ g ( z2 − z1 ) + ρ 

 2D 
 2D 3in.

d
ρ g ( z2 − z1 ) + ρ 
or
 fV 2 
 fV 2 
=




 D 3in.  D d
Using V =
f3in.Q3in.2
D3in.5
4Q
π D2
=
gives
fd Qd 2
Dd
5
or Dd 5 =
fd Qd 2
f3in.Q3in.2
D3in.5
(1)
The Reynolds number for the 3-in. pipe is

slugs  
ft 3 /s
gal   lb ⋅ s 2 

4 ( 0.89 )  1.94 3  
6.0



min   slug ⋅ ft 
ft   448.8gal/min  

ρVD 4 ρQ
=
=
= 1210
Re =
μ
π Dμ
in.2 
 3.068  
−7 lb ⋅ s  
ft   6.6 × 10
π
  144 2 
in.2  
ft 
 12

The flow is laminar so the Hagen-Poiseuille equation gives
f =
64 64ν 16π Dν
=
=
.
Re VD
Q
Assuming the flow is laminar for the pipe of diameter d , Eq. (1) gives
 16π Ddν  2

 Qd
Qd 
DQ

5
Dd =
D3in.5 = d d D3in.4 ,
Q3in.
 16π D3in.ν 
2

 Q3in.
 Q3in. 
Dd = 4
Qd
(10.7gal/min ) 3.068 in. ,
D3in. = 4
(
)
Q3in.
( 6.0 gal/min )
or
Dd = 3.55in.
1
The pipe size when choosing actual available pipe would be 3 in. schedule 40 pipe
2
(3.548 in. ID)
To verify that the flow is laminar in the pipe of diameter d , we calculate the Reynolds
number

slugs  
gal  
ft 3 /s

4 ( 0.89 )  1.94 3   10.7



min   449gal/min 
ft  

4 ρQ
Re =
=
π Dμ
lb ⋅ s  
in.2   slug ⋅ ft 
 3.548  
π
ft   6.6 × 10 −7 2   144 2  

ft  
ft   lb ⋅ s 2 
 12

Re = 1860
So the flow is laminar.
Problem 8.24
A 3-in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.) carries
210 °F SAE 40 crankcase oil at the rate of 6.0 gal/min . The oil specific gravity is 0.89 , and
the absolute viscosity is 6.6 ×10 −7 1b ⋅ s/ft 2. Calculate the pipe size required to carry the same
flowrate at approximately one-half the pressure drop of the 3-in. pipe. Both pipes are
horizontal.
Solution 8.24
Assume constant fluid density and apply the mechanical energy
equation from 1 to 2.
Δp
1
p1 V12
p V2
+
+ gz1 − ws = 2 + 2 + gz2 + ghL
ρ
2
ρ
2
ℓ
Now ws = 0 . The continuity equation gives V1 = V2 .
This gives
p1 − p2 = ρ ghL .
Denote the new (larger) pipe by the subscript d .
 f V2 
 f V2 
,
p
−
p
=
2
p
−
p
=
ρ
2
ρ
( 1 2 )3in. ( 1 2 )d



 ,
2
2
D
D

3in.

d
 fV 2 
 fV 2 

 = 2 

 D 3in.
 D d ,
or
Using V =
f3in.Q3in.2
D3in.5
4Q
π D2
=
,
2 fd Qd 2
Dd
5
or
Dd 5 =
2 fd Qd 2
f3in.Q3in.2
D3in.5
(1)
The Reynolds number in the 3-in. pipe is

slugs  
gal  
ft 3 /s

4 ( 0.89 )  1.94 3   6.0

 
min   449 gal/min 
ft  

ρVD 4 ρQ
Re3-in =
=
=
= 570.
μ
π Dμ
in.2   slug ⋅ ft 
 3.068  
−6 lb ⋅ s  
ft   1.4 × 10
π
  144 2  

in.2  
ft   lb ⋅ s 2 
 12

The flow is laminar so the Hagen-Poiseuille equation gives
f =
64 64ν 16π Dν
=
=
.
Re VD
Q
2
Assuming the flow is laminar in both cases, Eq. (1) gives
 16π Ddν 
2

Q

 D 5 = 2D D 4
5
Dd =
d 3in.
3in.
 16π D3in.ν 


Q


or
Dd = 4 2 D3in. = 4 2 ( 3.068 in.) = 3.648 in.
When choosing actual available pipe, the new pipe size is
1
3 in., schedule 40, commercial steel pipe, d = 3.548 in.
2
Note that the flow is laminar in the new pipe also since

slugs  
gal  
ft 3 /s

4 ( 0.89 )  1.94 3   6.0

 
min   449 gal/min 
ft  

4 ρQ
Re =
=
= 254
π Dν
in.2   slug ⋅ ft 
 3.548  
−6 lb ⋅ s  
ft   1.4 × 10
π
  144 2  

in.2  
ft   lb ⋅ s 2 
 12

so this flow is also laminar.
Problem 8.25
Water at 20 °C flows down a vertical pipe with no pressure drop. Find the range of pipe
diameters D (if any) for which the flow is definitely laminar.
•
1
Solution 8.25
V
Apply the mechanical energy equation from point 1 to point 2.
g
p1 V12
p2 V2 2
+
+ gz1 − ws =
+
+ gz2 + ghL
ρ
2
ρ
2
The problem statement gives p1 − p2 and ws = 0 .
•
Assuming constant water density, the continuity equation gives V1 −V2 . Let
z1 − z2 = h . This gives
LV2
L 8Q 2
8 f Q2
gh = ghL = f
= f
= 2 5
D 2
D π 2D 4
π D
(1)
The maximum Reynolds number for definitely laminar flow is about 2100 so
VD
ν
4Q
< 2100 or Q < 525 π Dν
π Dν
(2)
64 16 π Dν
.
=
Re
Q
(3)
=
and
f =
Equations (1) and (3) give
gh =
8 Q 2  16 π Dν  128 ν Q
=

=
π 2 D5 π 2D5  Q 
π D4
8 f Q2
or
D4 =
128 ν Q
.
π gh
Since the pipe is vertical, h =
D4 =
128ν Q
.
πg
and
(4)
2
Substituting for Q from Eq. (2) gives
D4 ≤
128ν
( 525π Dν )
πg
D≤3
67200ν 2
.
g
The numerical values give
D≤3
(
)
(9.807 m/s2 )
67200 1.0 × 10 −6 m 2 /s
D ≤ 0.00190 m = 1.90 mm
2
Problem 8.26
A person is donating blood. The pint bag in which the blood is collected is initially flat and is
at atmospheric pressure. Neglect the initial mass of air in the 1/8-in. I.D., 4 ft- long plastic
tube carrying blood to the bag. The average blood pressure in the vein is 40 mm Hg above
atmospheric pressure. Estimate the time required for the person to donate one pint of blood.
Assume that blood has a specific gravity of 1.06 and a viscosity of 1.0 ×10 −41b ⋅ s/ft 2. The
needle’s I.D. is 1/16 in. and the needle length is 2.0 in. The bag is 1.0 ft below the needle inlet
and the vein’s I.D. is 1/8 in. Optional: Donate a pint of blood and check your answer.
Solution 8.26
Assume steady flow conditions and apply the mechanical energy equation from
vein (point 1) to bag (point 2) inlet.
p1
ρ
+
α1V12
2
+ gz1 − ws =
p2
ρ
+
α 2V22
2
+ gz2 + ghL .
Now ws = 0 , use p2 = 0 psig , and z1 − z2 = 1 − 0 ft. The continuity equation gives V1 = V2
(same ID for vein and tube). Now assuming laminar flow,
V 2 64 V 2 64ν V 2
=
=
Re D 2
D 2
VD D 2
32ν V 32ν 4Q 128ν Q
=
= 2
=
π D4
D2
D π D2
ghL = f
The mechanical energy equation is (using h = z1 − z2 )
p1
ρ
+ gh = ( ghL )tube + ( ghL )needle =

128ν Q 



+ 4 
 4 

π  D tube  D  needle 
or
Q=
p1 + ρ gh

128μ 



+ 4 
 4 
π  D tube  D  needle 
The numerical values give

  14.696 lb   144 in.2 
atm
lb
p1 = ( 40 mm Hg ) 
 = 111 2 .

  2
2
ft
 760 mm Hg   in. atm  ft

and
 lb ⋅ s2 
lb
slug 
ft 

111 2 + 1.06 1.94 3  32.2 2  (1ft ) 
 ft ⋅ slug 
ft
ft 
s 



Q=




128 
4 ft
2in. 
−4 lb ⋅ s  
1
10
×
+

4
π 
ft 2    1 4  1
 
in.
  ft 

 
 16  
  96 
Q = 1.28 ×10 −4 ft3 /s.
The time to fill one pint bag is
 gal   ft 2   min 
(1 pint ) 



V
 8 pint   7.48 gal   60 s 
=
t=
3
Q

−4 ft 
 1.28 × 10

s 

or
t = 2.2min.
The actual time to donate one pint of blood is 5 to 10 min.
We calculate the Reynolds numbers to see if the flow is laminar.
3
slugs  

−4 ft 
4 (1.06 ) 1.94 3  1.28 ×10

s 
ft  

VD
4Q D ρ 4 ρQ
Re =
=
=
==
lb ⋅ s   slug ⋅ ft 
ν

π D 2 μ π Dμ
π D 1× 10 −4 2  

ft   lb ⋅ s 2 

3.35 ft
=
D
so
Retube =
3.35 ft
= 322
 1 
 ft 
 96 
and
Re needle =
3.35ft
= 643
 1

ft 

 16 ×12 
so the flows are both laminar.
DISCUSSION We will see in a later section of Chapter 8 that there is a mechanical energy
loss at the needle entrance and at the needle exit (into the tube). These are accounted for by

V 2 
 8Q 2
=  K needle + K needle 
ghL =  K needle + K needle 
4
 ent

 ent

exit  2
exit  π D


where D is the needle diameter. Including these terms in the energy loss term gives
Q = 1.0 ×10 −4 ft3 /s
and
t = 2.8 min .
Problem 8.27
For oil ( SG = 0.86 , μ = 0.025 Ns/m2) flow of 0.2 m3 /s through a round pipe with
diameter of 500 mm , determine the Reynolds number. Is the flow laminar or turbulent?
Solution 8.27
SG =
ρ
ρH O
= 0.86
2
(
)
ρoil = 0.86 ρH O = 0.86 ( 999 ) = 859
V=
2
kg
m3
Q
0.3
=
= 1.02 m/s
A π 0.5 2
( )
4
Re =
ρVD ( 859 )(1.02 )( 0.5 )
=
= 1.75 × 10 4
0.025
μ
Based on the criterion that Re < 2100 represents laminar flow, this flow is turbulent.
Problem 8.28
As shown in the figure below, the velocity profile for laminar flow in a pipe is quite
different from that for turbulent flow. With laminar flow, the velocity profile is parabolic;
with turbulent flow at Re = 10000, the velocity profile can be approximated by the powerlaw profile shown in the figure. (a) For laminar flow, determine at what radial location you
would place a Pitot tube if it is to measure the average velocity in the pipe. (b) Repeat part
(a) for turbulent flow with Re = 10000.
Turbulent with Re = 10000
r 1/5
u = 1 – __
__
1.0
Vc
[
R
]
r
__
R
Laminar with Re < 2100
u = 1 – __
r 2
__
(R)
Vc
0.5
u
r
R
Vc
0
0.5
u
__
1.0
Vc
Solution 8.28
For laminar or turbulent flow,
R
Q = AV = π R 2V =  udA =  u ( 2π rdr ) = 2π  urdr
r =0
a) Laminar flow:
  r 2 
 R2 R2 
R2
−
π R V = 2πVc  r 1 −   dr = 2πVc 
Vc
 =π
4 
2
 2
0 
  R  
R
2
V
1
u
Thus, V = Vc , for u = V = c , the equation for
gives
2
2
Vc
2
2
u 1
1
1
r
r
= = 1 −   , or   = Thus, r =
R = 0.707R
Vc 2
2
2
R
R
b) Turbulent flow:
R


π R2V = 2πVc  r 1 −
0
1
5
r
1
1
r 5
 r 
r
dr = 2π R2Vc    1 −  d  

R
R
R  R 
0 
r
r
r
Let y = 1 −   so that   = 1 − y and d   = −dy
R
R
R
Thus,
y =0
π R V = 2π R Vc  (1 − y )
2
2
1
5
y
( −dy )
y=1
6
1 1
5
5
= 2π R Vc  y − y  dy


0

2

5 5 
 25 
= 2π R2Vc  −  = 2π R2Vc  
 6 11
 66 
or V =
50
50
u
Vc For u = V = Vc , the equation for
gives
Vc
66
66
1
r
u 50  r  5
= 0.750 so that r = 0.750R
=
= 1 −  or
R
Vc 66  R 
Problem 8.29
Water at 10 C flows through a smooth 60-mm- diameter pipe with an average velocity of
m
8 . Would a layer of rust of height 0.005 mm on the pipe wall protrude through the
s
viscous sublayer? Justify your answer with appropriate calculations.
Solution 8.29
h = 0.005 mm = 5 × 10–6 m
1
DΔp
 τ 2
δS = sublayer thickness = ∗ , where u∗ =  w  and τ w =
4
u
 ρ 
fV 2 ρ
Δp
V2
=f
so that τ w =
Also,
D 2g
γ
8
 m
8
( 0.06 m )
VD  s 
=
= 3.67 ×105, we
For a smooth pipe with Re =
2
ν
m
1.307 ×10−6
s
obtain f = 0.0138 (see the figure below).
5v
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
ρ VD
Re = _____
μ
4
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
Thus,
2
kg 
 m 
0.0138  8   999.7 3 
m  = 110 N
 s  
τw =
8
m2
or
1
N 2

110 2 

m  = 0.332 m
u∗ = 
s
 999.7 kg 
3
m 

and

m2 
5 1.307 ×10 −6


s 
5ν

δs = ∗ =
= 1.97 × 10 −5 m > h=5 ×10 −6 m
m
u
0.332
s
Thus, the rust does not protrude through laminar sublayer.
Problem 8.31
ft
Water at 60 F flows through a 6-in.- diameter pipe with an average velocity of 15 .
s
Approximately what is the height of the largest roughness element allowed if this pipe is to
be classified as smooth?
Solution 8.31
Let h = roughness height. Thus, h = δ s , where δ s =
5ν
u∗
1
DΔp
1
 τ 2
. Since Δp = f
ρV 2 , we obtain
with u =  w  and τ w =
D2
4
 ρ 
ρ fV 2
f
τw =
or u∗ =
V
8
8
 ft   6 
15
ft
VD  s   12 
=
= 6.19 ×105 , we obtain
For a smooth pipe with Re =
2
ν
ft
1.21×10 −5
s
from the figure below f = 0.0125
∗
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
ρ VD
Re = _____
μ
4
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
1
ft
 0.0125  2  ft 
Thus, u = 
 15  = 0.593
s
 8   s
or

ft 2 
5 1.21×10 −5


s 
5ν

= 1.02 ×10−4 ft = 0.0012 in.
δs = ∗ =
ft
u
0.593
s
∗
Problem 8.32
Water is pumped between two tanks as shown in the figure below. The energy line is as
indicated. Is the fluid being pumped from A to B or B to A ? Explain. Which pipe has the
larger diameter: A to the pump or B to the pump? Explain.
Energy line
A
P
B
Solution 8.32
The energy line is horizontal if there is no head loss. With friction (a head loss) the
elevation of the energy line decreases in the direction of flow. Hence, the flow is from B
to A .
The pipe with the smallest diameter has the largest head loss per length

h
1 V2 
 i.e., L = f
.
D 2 g 

Thus, the pipe from A to the pump has the larger diameter.
Problem 8.33
A person with no experience in fluid mechanics wants to estimate the friction factor for
1-in.- diameter galvanized iron pipe at a Reynolds number of 8000. The person stumbles
64
across the simple equation of f =
and uses this to calculate the friction factor. Explain
Re
the problem with this approach and estimate their error.
Solution 8.33
For Re = 8000 under standard conditions, the pipe flow will be turbulent.
f - laminar
64
64
=
= 8 ×10 −3
Re 8000
f - turbulent
f =
For galvanized iron pipe, ε = 0.0005 ft so ,
ε
D
=
0.0005
= 6 ×10 −3
1
 
 
 12 
Making use of the Moody Chart
f ≈ 0.04
The error is in using the laminar equation to calculate the friction factor when the flow is
turbulent.
factual
f
0.04
= turbulent =
=5
flaminar
flaminar 0.008
That is, the friction factor is 5 times greater if the flow were laminar.
Problem 8.34
During a heavy rainstorm, water from a parking lot completely fills an 18-in.- diameter,
smooth, concrete storm sewer. If the flowrate is 10
ft3
, determine the pressure drop in a
s
100-ft horizontal section of the pipe. Repeat the problem if there is a 2-ft change in
elevation of the pipe per 100 ft of its length.
Solution 8.34
V2
ℓ
ft
= 100
(2)
V1
(1)
D = 18 m
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
D 2g
γ 2g
γ
2g
where p1 − p2 = Δp, γ = ρ g, and
In general,
(1)
ft3
Q
ft
s
= 5.66
V = V1 = V2 = =
2
A π  18 
s
ft


4  12 
Thus, Eq. (1) gives:
1
Δp = γ ( z2 − z1 ) + f
ρV 2
D2
ε 0.001ft
=
= 6.67 ×10 −4 (see the table below)
For a smooth concrete pipe
D  18 
 ft 
 12 
10
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
ft  18 

5.66  ft 

VD 
s  12 
=
= 7.02 ×105
and Re =
2
ν
ft
1.21×10−5
s
Thus, from the figure below f = 0.0185
(2)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
4
6
105
2(106)
8
106
4
6
2(107)
8
107
ρ VD
Re = _____
μ
a) With z1 = z2 Eq. (2) gives
slugs
(100 ft ) 1.94 3 
2
2
ft   5.66 ft  = 38.3 lb  1ft  = 0.266 psi

Δp = 0.0185




s
 18 
ft 2  144 in.2 

ft
(2)


 12 
b) With flow uphill z2 − z1 = 2 ft so that
 1ft 2 
lb 
lb

Δp =  62.4 3  ( 2ft ) 
+ 0.266 2 = 1.13 psi
2


ft 
in.

 144in. 
c) With flow downhill z2 − z1 = −2 ft so that
 1ft 2 
lb 
lb

Δp =  62.4 3  ( −2ft ) 
+ 0.266 2 = −0.601 psi
2


ft 
in.

 144in. 
4
6
8
0.00001
ε
__
D
Problem 8.35
Water flows through a horizontal plastic pipe with a diameter of 0.2 m at a velocity of
cm
10 . Determine the pressure drop per meter of pipe and the power lost to the friction per
s
meter of pipe.
Solution 8.35
The pressure drop in the pipe can be found from Δp = f
ρV 2
D 2
The friction factor is determined from the Moody chart.
ρVD ( 999 )( 0.1)( 0.2 )
Re =
=
= 1.8 × 10 4
3
−
μ
1.12 × 10
For plastic pipe, ε = 0.0 mm (a smooth pipe)
ε 0.0
=
= 0.0
D 0.2
From the Moody chart,
f = 0.026
So Δp per meter ( = 1 m)
2
 1   999 ( 0.1) 

Δp = (0.026) 

2
 0.2  

Δp = 0.649 Pa per meter of pipe
Power can be found from
P = ( Δp ) Q
m
2
π
Q = VA = ( 0.1)  ( 0.2 )  = 3.14 ×10 −3
s
4

−3
−3 N ⋅ m
P = ( 0.649 ) 3.14 ×10 = 2.04 ×10
= 2.04 ×10−3 W
s
3
(
)
Problem 8.36
Water flows downward through a vertical 10-mm- diameter galvanized iron pipe with an
m
and exits as a free jet. There is a small hole in the pipe 4 m above
average velocity of 5
s
the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe
m
through the hole? Repeat the problem if the average velocity is 0.5 .
s
Solution 8.36
D = 0.01 m
(1)
•
?
Q
4m=ℓ
(2)
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
, where p2 = 0, z2 = 0, z1 = 4m, V1 = V2 = V . Thus,
D 2g
γ 2g
γ
2g
1
V2
ρV 2 − γ
− z1 , or p1 = f
D2
D 2g
γ
With ε from the table below,
p1
=f
(1)
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
 m
5
( 0.01m )
ε 0.15mm
VD  s 
=
= 0.015 so that with Re =
=
= 4.46 ×10 4
2
ν
D 10mm
m
1.12 ×10 −6
s
we obtain f = 0.045 (see the figure below)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
107
ρ VD
Re = _____
μ
Thus, from Eq. (1),
2
kg   m 
N
 4m  1 
5 N
p1 = 0.045 
999
5
 

 − 9800 3 ( 4m ) = 1.86 × 10
3
 0.01m  2 
m  s 
m
m2
m
Since, p1 > 0, water will leak out of the pipe when V = 5
s
m
If V = 0.5 , then Re = 4.46 × 103 and f = 0.052
s
Thus, from Eq. (1),
2
kg  
m
N
 4m  1 
4 N
p1 = 0.052 
  999 3   0.5  − 9800 3 ( 4m ) = −3.66 × 10
s 
 0.01m  2 
m 
m
m2
m
Since p1 < 0, air will enter the pipe when V = 5
s
Note: The above conclusion is valid regardless of the length, .
4
6
8
0.00001
ε
__
D
Problem 8.37
Air at standard conditions flows through an 8-in.- diameter, 14.6-ft- long, straight duct
with the velocity versus head loss data indicated in the following table. Determine the
average friction factor over this range of data.
V (ft/min)
3950
3730
3610
3430
3280
3000
2700
h (in. water)
0.35
0.32
0.30
0.27
0.24
0.20
0.16
Solution 8.37
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
where V1 = V2 = V, Δp = p1 − p2 , z1 = z2
D 2g
γ 2g
γ
2g
1
2 ΔpD
ρV 2 or f =
Thus, Δp = f
where Δp = γ H2O h
D2
ρ V2
or
lb   8 
 h 
2  ft   62.4 3   ft 
ft
h
 12  
ft   12 
= 7.18 ×105 2 , where h ∼ in. of water, V ∼
f =
2
min
V
slugs 

 V ft 
 0.00238 3  (14.6 ft )  60 s 


ft 

Calculated values are given below:
V (ft/min)
h (in. water)
f
3950
3730
3610
3430
3280
3000
2700
0.35
0.32
0.30
0.27
0.24
0.20
0.16
Average f =
0.0161
0.0165
0.0165
0.0165
0.0160
0.0160
0.0158
0.0162
The average value of f is fave = 0.0162
Problem 8.38
Water flows through a horizontal 60-mm- diameter galvanized iron pipe at a rate of
0.02
m3
. If the pressure drop is 135 kPa per 10 m of pipe, do you think this pipe is (a) a
s
new pipe, (b) an old pipe with a somewhat increased roughness due to aging, or (c) a very
old pipe that is partially clogged by deposits? Justify your answer.
Solution 8.38
D = 0.06 m
(1)
V
ℓ = 10 m
p1 – p2 = 135 kPa
(2)
For the horizontal pipe, ( z1 = z2 ) with V1 = V2 , the energy equation
1
p1 V12
p V2
V2
ρV 2
+
+ z1 = 2 + 2 + z2 + f
reduces to p1 − p2 = f
D2
D 2g
γ 2g
γ
2g
or
2
N
10m 1 
kg  
m
135 × 103 2 = f
999
7.07


 , or f = 0.0324
0.06m 2 
s 
m
m3  
m3
0.2
m
Q
s
= 7.07
where we have used V = =
A π 0.06m 2
s
(
)
4
m

7.07  ( 0.06m )
VD 
ε 0.15 mm
s 
=
= 3.79 × 105 and
=
= 2.5 ×10 −3
with Re =
2
ν
D
60 mm
m
1.12 ×10−6
s
for a new galvanized iron pipe (see the table below),
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
The friction factor should be (see the figure below) f = 0.0255 .
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
4
6
2(107)
8
106
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Since this is less than the actual f = 0.324, the pipe is not a new pipe.
With Re = 3.79 ×105 and f = 0.324 , we obtain from the figure above a relative roughness of
ε
= 0.006 . This is approximately twice the roughness of a new pipe – certainly quite
D
possible. A very old partially clogged pipe would have considerably greater head loss. Thus,
the pipe is an old pipe with somewhat increased roughness.
Problem 8.39
Water flows at a rate of 10 gallons per minute in a new horizontal 0.75-in.- diameter
Δp
galvanized iron pipe. Determine the pressure gradient,
, along the pipe.
Solution 8.39
gal  1min   231in.3  1gal 
ft3
Q = 10

 = 0.0223


min  60 s   1gal   1728 in.3 
s
Thus,
ft3
0.223
Q
s = 7.27 ft
V= =
A π  0.75 2
s
ft 

4  12 
Now, for a horizontal pipe,
1
Δp = f
ρV 2 where since
D2
ft  0.75 
7.27 
ft 
VD
s  12 
Re =
=
= 3.76 ×10 4
2
ν
ft
1.21×10 −5
s
and
ε 0.0005ft
=
= 0.008
D  0.75 
ft 

 12 
it follows from the figure below that f = 0.037
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
104
6
2(105)
8
105
4
6
2(106)
8
4
106
ρ VD
Re = _____
μ
Thus,
2
slugs 
ft 

0.037 1.94 3   7.27 
lb  1ft 2 
psi
Δp
s
ft  

=
= 30.4 3 
=
0.211

ft
 0.75 
ft  144 in.2 
ft  (2)

 12 
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
Problem 8.40
Carbon dioxide at a temperature of 0 C and a pressure of 600 kPa (abs) flows through a
m
horizontal 40-mm- diameter pipe with an average velocity of 2 . Determine the friction
s
N
factor if the pressure drop is 235 2 per 10-m length of pipe.
m
Solution 8.40
For a horizontal pipe, Δp = f
1
2DΔp
ρV 2 , or f =
D2
ρ V2
where
N
kg
p
m2
ρ=
=
= 11.63 3
RT 
N⋅m 
m
188.9 kg ⋅ K  ( 273K )


600 ×103
Thus,
N 

2 ( 0.04 m )  235 2 

m  = 0.0404
f =
2
kg 

 m
11.63
10
m
2
(
)





m3 
 s 
Problem 8.41
Blood (assume μ = 4.5 ×10 −5
lb ⋅ s
ft 2
, SG = 1.0) flows through an artery in the neck of a giraffe
from its heart to its head at a rate of 2.5 ×10
−4 ft
3
s
. Assume the length is 10 ft and the
diameter is 0.20 in. If the pressure at the beginning of the artery (outlet of the heart) is
equivalent to 0.70 ft Hg , determine the pressure at the end of the artery when the head is
(a) 8 ft above the heart, or (b) 6 ft below the heart. Assume steady flow. How much of this
pressure difference is due to elevation effects, and how much is due to frictional effects?
Solution 8.41
(2)
V
ℓ = 10 ft
D = 0.20 in.
(1)
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
, where V1 = V2 = V
D 2g
γ 2g
γ
2g
and
ft3
Q
ρVD
s = 1.146 ft
V= =
Thus, Re =
, or
2
A π  0.2 
s
μ
ft 

4  12 
slugs 
ft  0.2 

ft 
1.94
1.146 
s 
s  12 

Re =
= 823. Hence, the flow is laminar with
−5 lb ⋅ s
4.5 ×10
ft 2
64 64
f =
=
= 0.0778
Re 823
lb 
lb

Also, p1 = γ Hg h =  847 3  ( 0.70 ft ) = 593 2
ft 
ft

2.5 ×10−4
Hence, from Eq. (1),
p2 = p1 − γ ( z2 − z1 ) − f
1
ρV 2
D2
(1)
(a) With z2 − z1 = 8 ft ,
lb 
10ft  1  
slugs  
ft 

1.94 3  1.146 
p2 = 593 2 −  62.4 3  ( 8ft ) − 0.0778



s
 0.2   2  
ft 
ft 
ft  
ft 

 12 
lb
= 593
lb
lb
lb
lb
ft
ft
ft
ft 2
− 499
2
Note: −499
lb
ft 2
− 59.5
2
= 34.5
2
2
is due to elevation, −59.5 is due to friction.
(b) With z2 − z1 = −6 ft ,
lb 
10 ft  1  
slugs  
ft 

p2 = 593 2 −  62.4 3  ( −6 ft ) − 0.0778
1.94 3  1.146 



s
 0.2   2  
ft 
ft 
ft  
ft 

 12 
lb
= 593
lb
ft
2
+ 374
Note: 374
lb
ft
2
lb
ft
2
− 59.5
lb
ft
2
= 908
lb
ft 2
is due to elevation, −59.5
lb
is due to friction.
ft
2
Problem 8.42
A 40-m- long, 12-mm- diameter pipe with a friction factor of 0.020 is used to siphon 30 C
water from a tank as shown in the figure below. Determine the maximum value of h
allowed if there is to be no cavitation within the hose. Neglect minor losses.
10 m
7m
3m
30 m
h
Solution 8.42
The minimum pressure is the vapor pressure pv = 4.243 kPa (abs) (see Table B.2 Physical
Properties of Water [SI Units]). Assume the minimum pressure is at the top of the hose
p3 = pv . We will check this assumption after we obtain h .
10 m
7m
(1)
•
•
(3)
3m
30 m
h
• (4)
L
Thus,
p V2
p1 V12
V2
+
+ z1 = 3 + 3 + z3 + f
, where p1 = 101 kPa(abs), z1 = 3m ,
D 2g
γ 2g
γ 2g
z3 = 7m, V1 = 0, V3 = V , and p3 = 4.243 kPa (abs) . Thus, with f = 0.02
kN
V2
m2 + 3m = 7m + 1 + 0.02  10 m  



kN
 0.012 m   2  9.81 m 

9.77 3


m
s2 

(101 − 4.243)
or
m
s
Obtain h from
V = 2.56
m
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
, where p2 = 0, V2 = V = 2.56 ,
D 2g
γ 2g
γ
2g
s
z2 = −h , and
= 40 m . That is, with p1 = p2 = 0
2
m

2.56 


 40 m   
s 
3m = − h + 1 + 0.02 
, or hmax = 19.6 m

 0.012 m   2  9.81 m 



s2 

Check if minimum pressure occurs at (3). Consider point (4).
V4 2
p2 V2 2
LV2
+
+ z4 =
+
+ z2 + f
with p2 = 0, V2 = V4 = V
From
D 2g
γ
2g
γ
2g
p4
(3)
30 m
(4)
•
h + 7 = 26.6 m
Z4
L
(2)
we obtain p4 = γ ( z2 − z4 ) + f
L1
ρV 2
D2
If we use z2 = 0 , then from the figure:
30
L
=
, or L = 1.128z4
z4 26.6
Thus,
p4 = 9.80

− z 4 ) + ( 0.02 ) 
3 (
kN
m
1.128 z 4   1  
kg  
m
    999 3   2.56 
s 
m 
 0.012   2  
2
or
(
)
p4 = −9.80 × 103 + 6.15 × 103 z4 = −3650 z4
Thus, p4 decreases as z4 increases. That is, the minimum pressure occurs at section (3) as
assumed.
Problem 8.43
m3
Gasoline flows in a smooth pipe of 40-mm diameter at a rate of 0.001
. If it were
s
possible to prevent turbulence from occurring, what would be the ratio of the head loss for
the actual turbulent flow compared to that if it were laminar flow?
Solution 8.43
Let ( )t denote the turbulent flow and ( ) the laminar flow.
V2
V2
Thus, hLt = ft
and hL = f
D 2g
D 2g
where
m3
0.001
Q
s = 0.796 m
V = Vt = V = =
s
A π 0.04m 2
(
)
4
(1)
From Table 1.5 Approximate Physical Properties of Some Common Liquids (SI Units)
ρ = 680
kg
m
3
and μ = 3.1× 10 −4
N ⋅s
m2
so that
kg 
m

680 3  0.796  ( 0.04m )

ρVD 
s 
m 
Re =
=
= 6.98 ×10−4
N
s
⋅
μ
3.1×10−4 2
m
Hence, from the figure below, for a smooth pipe ft = 0.0192
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
ρ VD
Re = _____
μ
4
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
while for laminar flow f =
64
64
=
= 9.16 ×10 −4
4
Re 6.98 ×10
Thus, from Eq. (1)
hLt
hL
=
ft
0.192
=
= 21.0
f
9.16 ×10 −4
Problem 8.44
A 3-ft- diameter duct is used to carry ventilating air into a vehicular tunnel at a rate of
9000 ft3 /min . Tests show that the pressure drop is 1.5 in. of water per 1500 ft of duct. What
is the value of the friction factor for this duct and the approximate size of the equivalent
roughness of the surface of the duct?
Solution 8.44
p1 V12
p2 V2 2
V2
+
+z =
+
+ z2 + f
,
D 2g
γ 2g 1 γ
2g
lb  1.5 
lb

ft  = 7.80 2
where z1 = z2 , V1 = V2 , and p1 − p2 = γ H2O h =  62.4 3  
ft   12 
ft


ft3   1min 
 9000


min   60s 
Q 
ft
= 21.2
Also, V = =
π
A
s
(3ft )2
4
Thus, from Eq. (1)
p1 − p2 = f
1
ρV 2 or
D2
lb 

2 ( 3ft )  7.80 2 
2D ( p1 − p2 )
ft 

=
= 0.0292
f =
2
2
ρ V
ft 


−3 slug 
 2.38 ×10
 (1500 ft )  21.2 s 


ft3 

ft 

21.2  ( 3ft )

VD 
s
=
= 4.05 ×105
From Fig. 8.20 with f = 0.0292 and Re =
2
ν
ft
1.57 ×10 −4
s
we obtain
ε
D
= 0.0044 . Thus, ε = 0.0044 ( 3 ft ) = 0.0132 ft
(1)
Problem 8.45
H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation
f smooth ≈ 0.3164 Re −1 4 is reasonably accurate for Reynolds numbers between 4000 and 105.
Use this information for the following scenario. Water at 20 °C is to flow through a 3-cm
I.D. plastic pipe at the rate of 0.001 m 3 / s . Find the incline angle of the pipe needed to
make the static pressure constant along the pipe.
Solution 8.45
Apply the mechanical energy equation from 1 to 2 .
1•
ϕ
p1 V12
p V2
+
+ gz1 + ghp = 2 + 2 + gz2 + ght + ghL .
ρ
ρ
2
2
ℓ
Now the problem statement gives p1 = p2 and ghp = ght = 0 and
the continuity equation gives V1 = V2 .
The energy equation is
g ( z1 − z2 ) = ghL
where
ghL = f
V2
and ( z1 − z2 ) = sinθ
D 2
so
g sin θ = f
V2
D 2
or sin θ =
fV 2
.
2 gD
Since plastic is a smooth pipe, the given Blasius equation gives
1
1
−
VD
4


f = 0.3164 Re 4 = 0.3164
−


 ν 
so
sin θ = 0.1582
V 1.75ν 0.25
gD1.25
.
The numerical values give

m3 
4  0.001


s
Q 4Q
 = 1.41 m ,
= 
V= =
2
2
A πD
s
π ( 0.03 m )
2

−6 m 
1.0
10
×


s 

m

1.25
 9.81 2  ( 0.03)
s 

1.75
sinθ = 0.1582
m

1.41 
s 

0.25
= 0.0774,
•2
and
θ = 4.44; inclined downward in the direction of flow.
Checking the Reynolds number gives
m

 m 
1.41  ( 3.0 cm ) 


VD 
s 
 100 cm  = 42300
Re =
=
2
ν

−6 m 
1.0 ×10

s 

so the flow is turbulent and in the proper range for the given Blasius equation.
Problem 8.46
Von Karman suggested that the wholly turbulent friction factor be expressed by the
equation
1
f =
2

 ε 
4 0.57 − log  
 D 

where ε is the absolute roughness of the pipe. Compare the values predicted by this
equation and those indicated on the Moody chart.
Solution 8.46
Choosing six values of the relative roughness,
ε
D
0.0001
0.0004
0.001
0.004
0.01
0.04
f
(Von Karman)
0.0120
0.0159
0.0196
0.0284
0.0379
0.0646
ε
D
, gives the following:
f
(Moody Chart)
0.013
0.016
0.0195
0.0283
0.038
0.065
Problem 8.47
The Swamee and Jain formula for the friction factor is
0.25
f =
.
2
  ε
5.74  
 log  3.7D + 0.9  
Re  
 
ε
= 0.00001 , 0.0001 , 0.001 , and 0.01 and Reynolds numbers
D
of 10 4 , 105 , 106 , and 10 7 with the Moody chart and decide whether it is an acceptable
replacement for the Colebrook formula.
Compare this equation for
Solution 8.47
f =
0.25
(1)
2
  ε
5.74  
 log  3.7D + 0.9  
Re  
 
Tabulate the friction factor f from both the Moody chart and Eq. (1) for the above values
of
ε
D
and Re .
Percent Diff
ε/D
0.00001
0.0001
0.001
0.01
Re
Moody f
Eq. (1)
104
0.0313
0.0310
0.96%
105
0.0182
0.0179
1.65%
6
10
0.0119
0.0119
0.00%
107
0.0089
0.0091
2.25%
4
10
0.0313
0.0311
0.64%
105
0.0183
0.0185
1.09%
106
0.0133
0.0135
1.50%
7
10
0.0120
0.0122
1.67%
104
0.0330
0.0327
0.91%
5
10
0.0225
0.0223
0.89%
106
0.0200
0.0200
0.00%
107
0.0196
0.0197
0.51%
4
10
0.0437
0.0440
0.69%
105
0.0385
0.0388
0.78%
6
10
0.0380
0.0380
0.00%
107
0.0380
0.0379
0.26%
1−
Eq. (1)
× 1000
Moody f
We conclude
Swamee and Jain formula is an acceptable substitute for Colebrook formula for
0.00001 ≤
ε
D
≤ 0.01 and 10 4 ≤ Re ≤ 107.
Problem 8.48
The Haaland formula for the friction factor is
0.3086
.
f =
2
1.11

 6.9
 ε   
+ 
log 
 
 3.7D   
 Re

ε
= 0.00001 , 0.0001 , 0.001 , and 0.01 and Reynolds
D
numbers of 10 4 , 105 , 106 , and 10 7 with the Moody chart and decide whether it is an
acceptable replacement for the Colebrook formula.
Compare this equation for f for
Solution 8.48
0.3086
f =
(1)
2
1.11  

 6.9

 ε 
+ 
log 
 
 3.7D   

 Re
We will tabulate the friction factor from both the Moody chart and Eq. (1) for the above
values of
ε/D
ε
D
and Re .
Re
Moody f
Percent Diff
Eq. (1)
1−
0.00001
0.0001
0.001
0.01
Eq. (1)
× 1000
Moody f
104
0.0313
0.0309
1.28%
105
0.0182
0.0179
1.65%
6
10
0.0119
0.0118
0.84%
107
0.0089
0.0090
1.12%
104
0.0313
0.0310
0.96%
5
10
0.0183
0.0183
0.00%
106
0.0133
0.0133
0.00%
7
10
0.0120
0.0122
1.67%
104
0.0330
0.0322
2.42%
105
0.0225
0.0220
2.22%
6
10
0.0200
0.0199
0.05%
107
0.0196
0.0197
0.51%
4
10
0.0437
0.0430
1.60%
105
0.0385
0.0385
0.00%
106
0.0380
0.0380
0.00%
7
0.0380
0.0380
0.00%
10
We conclude
Haaland formula is an acceptable substitute for the Colebrook formula for
0.00001 ≤
ε
D
≤ 0.01 and 104 ≤ Re ≤ 107.
Problem 8.49
The Churchill formula for the friction factor is
1
12

 8 
1

f = 8 
 +
1.5
 Re 
( A + B ) 
12
,
where
16

 7 0.9
ε  
+
A = −2.457 ln 
 ;

Re
3.7
D



 


16
 37530 
B =
 .
 Re 
Compare this equation for f for both the laminar and turbulent regimes for
ε
= 0.00001, 0.0001, 0.001, and 0.01, and Reynolds numbers of
D
10, 102 , 103 , 10 4 , 105 , 106 , and 107 with the Moody chart and decide whether it is an
acceptable replacement for the Colebrook formula.
Solution 8.49
We will tabulate the friction factor from both the Moody chart and the Churchill formula
for the above values of
ε
and Re . This is done below and the comparison of the tabulated
D
values gives the conclusion:
The Churchill formula is an acceptable substitute for the Colebrook formula for
0.00001 ≤
ε
D
≤ 0.01 and 10 ≤ Re ≤ 107.
Percent Diff
ε /D
Re
Moody f Church. f
0.00001
10
10²
10³
104
105
106
107
10
102
103
104
105
106
107
10
102
103
104
105
106
107
10
102
103
104
105
106
107
6.40
0.640
0.0640
0.0313
0.0182
0.0119
0.0089
6.40
0.640
0.0640
0.0313
0.0183
0.0133
0.0120
6.40
0.640
0.0640
0.0330
0.0225
0.0200
0.0196
6.40
0.640
0.0640
0.0437
0.0385
0.0380
0.0380
0.0001
0.001
0.01
6.40
0.640
0.0640
0.0310
0.0179
0.0119
0.0091
6.40
0.640
0.0640
0.0312
0.0185
0.0135
0.0122
6.40
0.640
0.0640
0.0327
0.0223
0.0220
0.0197
6.40
0.640
0.0640
0.0441
0.0387
0.0380
0.0379
1−
Church. f
× 100
Moody f
0.00%
0.00%
0.00%
0.96%
0.16%
0.00%
2.25%
0.00%
0.00%
0.00%
0.32%
1.09%
1.48%
1.67%
0.00%
0.00%
0.00%
0.91%
0.89%
0.00%
0.51%
0.00%
0.00%
0.00%
0.92%
0.52%
0.00%
0.26%
Problem 8.50
Air at standard temperature and pressure flows through a 1-in.- diameter galvanized iron
ft
pipe with an average velocity of 8 . What length of pipe produces a head loss equivalent
s
to (a) a flanged 90 elbow, (b) a wide-open angle valve, or (c) a sharp-edged entrance?
Solution 8.50
 ft  1 
8
ft 
VD  s 
KLD
12 

=
= 4.25 ×103. Thus, with
where with Re =
eq =
2
ν
f
ft
1.57 ×10−4
s
ε 0.0005ft
=
= 0.006 (see the table below)
D  1 
 ft 
 12 
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
we obtain f = 0.045 (see the figure below)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
4
6
2(107)
8
106
4
6
107
ρ VD
Re = _____
μ
1 
K L  ft 
 12  = 1.852 K or a) 90 elbow: K = 0.3 or
Thus, eq =
L
L
0.045
b) globe valve: K L = 2 or
eq = 0.556 ft
eq = 3.70 ft
c) sharp entrance: K L = 0.5 or
eq = 0.926 ft
8
0.00001
ε
__
D
Problem 8.51
Given 90 threaded elbows used in conjunction with copper pipe (drawn tubing) of
0.75-in. diameter, convert the loss for a single elbow to equivalent length of copper pipe for
wholly turbulent flow.
Solution 8.51
eq =
KLD
f
For 90 -threaded elbow, K L = 1.5
For copper pipe (drawn tubing), ε = 0.000005 ft
So
ε
D
=
0.000005
= 8 ×10 −5
0.75
12
From Moody chart (wholly turbulent flow),
f ≅ 0.0115
eq =
(1.5 ) 
0.75 

 12  = 8.15 ft
0.0115
Problem 8.52
To conserve water and energy, a “flow reducer” is installed in the shower head as shown in
the figure below. If the pressure at point (1) remains constant and all losses except for that
in the flow reducer are neglected, determine the value of the loss coefficient (based on the
velocity in the pipe) of the flow reducer if its presence is to reduce the flowrate by a factor
of 2 . Neglect gravity.
1
__
in.
2
Flow reducer washer
(1)
Q
50 holes of
diameter 0.05 in.
Solution 8.52
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 , where p2 = 0, z1 = z2
2g
γ 2g
γ
Without the reducer
and
V1 =
Q
4Q
=
=
A1 π D12
V2 =
Q
=
A2
4Q
 0.5 
π
ft 
 12 
2
= 733Q

Q
=
1467
V1 and V2
2
 0.05 

50π 
ft 
12


4Q
ft
, and Q
s
ft 3 

s 
Thus,
p1 =
1
1
lb
ρ (V2 2 −V12 ) = ρ (14672 Q 2 − 733Q 2 ) = 8.07 × 105 ρQ 2 2 ,
2
2
ft
ft
s
ft
With the flow reducer, the flowrate is reduced by factor of two.
where ρ ∼
slug
(1)
3
, Q∼
1
1
Thus, V1 = (733Q ) and V2 = (1467Q ) with
2
2
(2)
1
p1 V12 V22
V2
+
=
+ K L 1 or p1 = ρ (V22 + ( K L − 1)V12 )
2
γ 2g 2g
2g
(3)
Hence, by combing Eqs. (1), (2), and (3), we obtain
2
2
1  1467 
 733  
Q  + ( K L − 1) 
Q 
8.07 × 10 ρQ = ρ 
2  2

 2
 
5
or
K L = 9.00
2
Problem 8.53
m3
in a 0.12-m- diameter pipe that contains a sudden
Water flows at a rate of 0.040
s
contraction to a 0.06-m- diameter pipe. Determine the pressure drop across the contraction
section. How much of this pressure difference is due to losses and how much is due to
kinetic energy changes?
Solution 8.53
D1 = 0.12 m
D2 = 0.06 m
•
m3
(1) Q = 0.04 s
• (2)
p1 V12
p2 V2 2
V2 2
+
+z =
+
+ z2 + K L
, where z1 = z2
γ 2g 1 γ
2g
2g
and
V1 =
0.04
m3
s
0.04
m3
s
Q
m
Q
m
=
= 3.54 , V2 =
=
= 14.1
A1 π (0.12 m)2
A2 π (0.06 m)2
s
s
4
4
2
2
 0.06 m 
A D 
Thus, with 2 =  2  = 
 = 0.25 ,
A1  D1 
 0.12 m 
we obtain from the figure below K L = 0.40
1.0
b
b
Separated flow
0.8
a
D
a
Secondary
flow
ℛ
0.6
90°
Primary
flow
0.4
ε
__ = 0.01
D
0.002
0.2
0.001
0
0
0
2
4
6
ℛ/D
8
10
12
Hence, from Eq. (1),
2
2
2
1 
kg  
m 
m 
m 

2
2
2 1
p1 − p2 = ρ K LV2 +V2 −V1 =  999 3   0.40 14.1  +  14.1  −  3.54  
 2
2 
s  
s  
s  
m  

or
p1 − p2 = 39.7 × 103
N
m
2
+ 93.0 ×103
N
m2
= 133 kPa
This represents a 39.7 kPa drop from losses and a 93.0 kPa drop due to an increase in
kinetic energy.
Problem 8.54
Water flows from the container shown in the figure below. Determine the loss coefficient
needed in the valve if the water is to “bubble up” 3 in. above the outlet pipe. The entrance
is slightly rounded.
Vent
27 in.
1 -in.-diameter galvanized iron pipe
2
with threaded fittings
18 in.
3 in.
2 in.
32 in.
Solution 8.54
Vent
(1)
•
27 in.
1 -in.-diameter galvanized iron pipe
2
with threaded fittings
(3)
•
18 in.
3 in.
(2) •
2 in.
32 in.
Determine the outlet velocity from Bernoulli equation:
p3 V32
V2 2
+
+ z2 =
+
+ z3 , where z3 − z2 = 3 in. , p2 = p3 = 0 , and V3 = 0
γ
2g
γ 2g
p2
Thus,
ft   3 
ft

V2 = 2 g ( z3 − z2 ) = 2  32.2   ft  = 4.01
s   12 
s

Also,
2
p3 V32
p1 V12

V
+
+z =
+
+ z + f +  KL 
, where p1 = p3 = 0 ,V1 = V3 = 0 ,
γ 2 g 1 γ 2 g 3  D
 2g
z1 =
ft
(18 + 27)
(2 + 3)
ft = 3.75ft , z3 =
ft = 0.417ft , and V = 4.01
12
12
s
Thus,
2

V
z1 = z3 +  f +  K L 
, where  K L = K Lentrance + 2 K Lelbow + K Lvalve
 D
 2g
or  K L = 0.2 + 2(1.5) + K Lvalve = 3.2 + K Lvalve
Pipe
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
Equivalent Roughness, ε
Feet
Millimeters
0.003–0.03
0.9–9.0
0.001–0.01
0.3–3.0
0.0006–0.003 0.18–0.9
0.00085
0.26
0.0005
0.15
0.00015
0.045
0.000005
0.0015
0.0 (smooth)
0.0 (smooth)



ε
0.0005ft 
 = 0.012, and
From the table above,
=
D   0.5  
  12 ft  


ft  0.5

4.01 
ft

VD 
s  12
Re =
=
= 13800
2
ν
ft
1.21×10 −5
s
Thus, from the figure below, f = 0.043
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
107
ρ VD
Re = _____
μ
Hence,
2
ft 

4.01 



s
 18 + 32 + 2 
3.75 ft = 0.417 ft +  0.043 
 + 3.2 + K Lvalve 
0.5



 2  32.2 ft 


s2 

or
K Lvalve = 5.68
4
6
8
0.00001
ε
__
D
Problem 8.55
New hi-tech fountains Ancient Egyptians used fountains in their palaces for decorative and
cooling purposes. Current use of fountains continues but with a hi-tech flair. Although the
basic fountain still consists of a typical pipe system (i.e., pump, pipe, regulating valve, nozzle,
filter, and basin), recent use of computer-controlled devices has led to the design of
innovative fountains with special effects. For example, using several rows of multiple nozzles,
it is possible to program and activate control valves to produce water jets that resemble
symbols, letters, or the time of day. Other fountains use specially designed nozzles to produce
coherent, laminar streams of water that look like glass rods flying through the air. Using fastacting control valves in a synchronized manner it is possible to produce mesmerizing threedimensional patterns of water droplets. The possibilities are nearly limitless. With the initial
artistic design of the fountain established, the initial engineering design (i.e., the capacity and
pressure requirements of the nozzles and the size of the pipes and pumps) can be carried out.
It is often necessary to modify the artistic and/or engineering aspects of the design in order to
obtain a functional, pleasing fountain. (See Problem 8.55.)
The fountain shown in the figure below is designed to provide a stream of water that rises
h = 10 ft to h = 20 ft above the nozzle exit in a periodic fashion. To do this, the water from
the pool enters a pump, passes through a pressure regulator that maintains a constant
pressure ahead of the flow control valve. The valve is electronically adjusted to provide the
desired water height. With h = 10 ft , the loss coefficient for the valve is K L = 50 . Determine
the valve loss coefficient needed for h = 20 ft . All losses except for the flow control valve
are negligible. The area of the pipe is 5 times the area of the exit nozzle.
h
4 ft
Pump
Flow control valve
Pressure regulator
Solution 8.55
For any height h ,
p1
γ
+ z1 +
V12
p
V2
− hL = 2 + z2 + 2 , where z1 = 0,
2g
γ
2g
z2 = h + 4ft, p2 = 0, V2 = 0 , and
V12
hL = K L
2
Thus,
p1 V12
V12
+
− KL
= z2
2g
γ 2g
(1)
For h = 10ft : ( K L = 50)
p1
γ
= z2 −
V12
V2
V2
+ K L 1 = (10ft + 4ft) + (50 − 1) 1
2g
2g
2g
(2)
Also, from (3) to (2):
p3
γ
+ z3 +
V32 p2
V2
=
+ z2 + 2 , where p2 = p3 = 0 , z2 − z3 = h , and V2 = 0
2g
γ
2g
Thus,
V32
= h and V3 = 2 gh
2g
(3)
ft 
ft

so for h = 10 ft , V3 = 2  32.2 2  (10 ft ) = 25.4
s
s 

A
ft 
ft
 1 
Also, V1A1 = V3A3 so that V1 = 3 V3 =    25.4  = 5.08
s
s
A1
 5 
Hence, Eq. (2) gives
2
ft 

14ft + 49  5.08 
p1
s

=
= 33.6 ft
γ
 
ft  
 2  32.2 2  
s 
 
For h = 20 ft , from Eq. (3): V3 =
( 2 )  32.2 2  ( 20 ft ) = 35.9
ft

s 
ft
s
Hence,
V1 =
A3
ft 
ft
 1 
V3 =    35.9  = 7.18
A1
s
s
 5 
p1
= 33.6 ft obtained above for
γ
h = 10 ft is also valid for h = 20 ft . Thus, with z2 = h + 4ft = 20ft + 4ft = 24 ft, Eq. (1) is
Since p1 is constant (independent of h ), the value
2
2
ft 
ft 


K L  7.18 
33.6 ft +  7.18 
s
s


−
= 24 ft
 
 
ft  
ft  
 2  32.2 2  
 2  32.2 2  
s 
s 
 
 
or
KL =13.0
Problem 8.56
Water flows through the screen in the pipe shown in the figure below as indicated.
Determine the loss coefficient for the screen.
V = 20 ft/s
Water
Screen
6 in.
SG = 3.2
Solution 8.56
V = 20 ft/s
(1)
•
Screen
(2)
•
Water
ℓ
6 in.
SG = 3.2
ft
p1 V12
p2 V22
V2
+
+ z1 =
+
+ z2 + K L
, where z1 = z2 ,V1 = V2 = V = 20
s
γ 2g
γ
2g
2g
Thus, K L =
2(p1 − p2 )
ρV 2
where p1 + γ = p2 + γ ( − 6 in.) + SG γ (6 in.)
or p1 − p2 = γ (SG − 1)(6 in.)
Hence,
lb 

6 
2  62.4 3  ( 3.2 − 1)  ft 
 12  = 0.177
ft 
KL = 
2
slugs  ft 

1.94 3   20 s 

ft  

Problem 8.57
Air flows though the mitered bend shown in the figure below at a rate of 5 cfs . To help
straighten, the flow after the bend, a set of 0.25-in.- diameter drinking straws is placed in the
pipe as shown. Estimate the extra pressure drop between points (1) and (2) caused by these
straws.
(1)
Tightly packed 0.25-in.-diameter,
12-in.-long straws
(2)
12 in.
Solution 8.57
The extra pressure drop, Δp, is equal to the pressure drop through the length of the straws
minus the pressure drop in that 12 in. length of the pipe without the straws. That is
ft3
5
Q
ft
1
s
= 6.37
ρV 2 with V = =
Δp = Δps − Δpns , where Δpns = f
2
A π  12 
s
D2
 ft 
4  12 
ft 

6.37  (1ft )

VD 
s
=
= 4.06 ×10 4. If we assume the pipe is smooth, it follows from
Also, Re =
2
ν
ft
1.57 ×10−4
s
the figure below that f = 0.0215.
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
6
2(104)
8
103
4
104
6
2(105)
8
4
6
105
2(106)
8
4
6
106
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Thus,
2
ft 
 12in.   1  
−3 slug  
−3 lb
Δpns = 0.0215 
    2.38 × 10
 6.37  = 1.04 × 10

3
s
ft  
ft 2
 12in.   2  
(1)
1
ρV 2 where the values of f , D , and V are different
D2
than those used above. In general, the flow geometry is quite complex – flow through the
straws and flow in the gaps between the straws. For simplicity, assume the gaps act as a
3
3
circular flow area of diameter Dg = D = ( 0.25in.) = 0.0938in.
8
8
With the straws in place, Δpns = f
0.25 in.
Dg
straw
ε
__
D
Thus, in each 0.5in. by 0.5in. cross section, there are four straws, or a total of
straws  π
N =4
(12 in.)2  , that is, N = 1810 straws.
2 

0.25 in.  4
(
)
If the flow is laminar, then Q  D4 so that
4
 0.0938 in. 
=
 = 0.0198
Qstraws  0.25 in. 
Qgap
That is, only about 2% of the flow is in the gap region – neglect this amount.
ft 3
Q
ft
s
Thus, V =
=
= 8.10
2
NA
s
 π  0.25  
1810  
ft  
 4  12  


5
Hence,
ft  0.25 

8.10 
ft 

VD 
s  12 
Re =
=
= 1070 < 2100 , the flow is laminar with
2
ν
−4 ft
1.57 ×10
s
f =
 12 in.   1  
ft 
64
64
−3 slug  
8.10
=
= 0.0598, or Δps = 0.0598 
    2.38 ×10



s
Re 1070
ft3  
 0.25 in.   2  
or Δps = 0.224
lb
ft 2
. Hence, when combined with result (1)
Δp = Δps − Δpns = (0.224 − 0.00104)
lb
ft
2
= 0.223
lb
ft 2
2
Problem 8.58
As shown in the figure below, water flows from one tank to another through a short pipe
whose length is n times the pipe diameter. Head losses occur in the pipe and at the entrance
and exit. Determine the maximum value of n if the major loss is to be no more than 10% of
the minor loss and the friction factor is 0.02 .
D
𝓵 = nD
Solution 8.58
If hLmajor = 10 0 0 hLminor , then
10 f
V2
V2
K
=  KL
or
= L
D 2g
D 10 f
2g
where  K L = K Lentrance + K Lexit = 0.8 + 1 = 1.8
Thus, with f = 0.2 and
1.8
nD
=
D 10(0.02)
or
n=9
= nD , Eq. (1) becomes
(1)
Problem 8.59
Water flows steadily through the 0.75-in. -diameter galvanized iron pipe system shown in
the figure below at a rate of 0.020 cfs . Your boss suggests that friction losses in the straight
pipe sections are negligible compared to losses in the threaded elbows and fittings of the
system. Do you agree or disagree with your boss? Support your answer with appropriate
calculations.
6-in. length
90° threaded
elbows
6-in. length
0.60-in. dia.
Reducer
1-in. length
4-in. length
Q = 0.020 cfs
Tee
Closed ball
valve
Solution 8.59
Major loss = f
V2
where
D 2g
= (6 + 6 + 4 + 1)in. = 17 in. , D = 0.75in.
and
ft3
ft
Q
s
= 6.52
V= =
2
s
A π  0.75  2
ft


4  12 
0.02
Thus, with
ft  0.75 
ft 

VD
s  12 
Re =
=
= 3.37 ×10 4 and
2
ν
ft
1.21×10 −5
s
ε 0.0005ft
=8 ×10 –3 (see the table below)
=
D  0.75 
ft 

 12 
6.52
Equivalent Roughness, ε
Pipe
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
Feet
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
Millimeters
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
we obtain (see the figure below) f = 0.038
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
V2
V2
17in. V 2
= 0.038
= 0.861
D 2g
0.75in. 2 g
2g
6
2(106)
8
106
ρ VD
Re = _____
μ
so that f
4
4
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
Also,
Minor loss= K L
V2
V2
V2
= [ 2(1.5) + 2 + 0.15]
= 5.15
2g
2g
2g
2
90°elbow
tee
A  0.6 in. 
reducer with 2 = 
 = 0.64 (see the figure
A1  0.75 in. 
below)
0.6
A1
A2
V22
hL = KL ___
2g
0.4
KL
0.2
0
0
0.2
0.4
0.6
0.8
1.0
A2/A1
Thus, from Eqs. (1) and (2):
V2
0.861
major loss
2g
=
= 0.167 = 16.7%
minor loss
V2
5.15
2g
Probably disagree with boss because pipe friction is about 17% of other losses.
Problem 8.60
Given two rectangular ducts with equal cross-sectional area, but different aspect ratios
(width/height) of 2 and 4 , which will have the greater frictional losses? Explain your
answer.
Solution 8.60
The duct with the greater losses is the one with the largest head loss per length,
Q
V2
hL = f
. If the areas are equal, then the velocities are equal since V = .
A
Dh 2 g
Let ( )2 and ( )4 denote ducts with aspect ratios of 2 and 4, respectively.
Thus,
f4 V4 2
f2 V2 2
 hL 
 hL 
=
=
and
, where V2 = V4 .
 
 
 4 Dh4 2 g
 2 Dh2 2 g
Hence,
f4
 hL 
 
f Dh
  4 Dh4
=
= 4 2
f2
f2 Dh4
 hL 
 
 2 Dh2
(1)
a
Let A2 = (2a )a
2a
a4
4a4
and A4 = (4a4 )a4
Thus, since A2 = A4
2a 2 = 4 a42 , or a4 =
1
a
2
and
Dh2 =
4 A2
4(2a 2 )
4
=
= a = 1.33a
P2
[ 4a + 2a ] 3
(2)
and
Dh4 =
4 A4
4(2a 2 )
4 2
=
=
a = 1.13a
8 
P4
5
 2
 2 a + 2 a


(3)
hL
, where
so that
4
a
5
3
=
=
= 1.179 so that Eq. (1) becomes
Dh4 4 2
3 2
a
5
Dh2
 hL 


f

4
= 1.179 4
f2
 hL 



2
(4)
ε
ε 

In general, f = f  Re,  in such a way that if
increases, f increases, and if Re
D
D

decreases, f increases. This is seen from the Moody chart as indicated below.
f
laminar
(4)
•
turbulent
•
(2)
(4)
•
ε /Dh
•
(2)
ε /Dh
4
2
Reh
 ε   ε 
For a given ε , 
 >
 since Dh4 < Dh2 [see Eqs. (2) and (3)].
D
D
 h  4  h 2
Also, since Reh =
VDh
ν
, it follows that
Re h4 < Re h2 since Dh4 < Dh2 and V2 = V4 .
Thus, whether the flow is laminar or turbulent, it follows that f4 > f2 . If follows from Eq.
(4) that
 hL 



4
>1
 hL 



2
That is, the duct with the aspect ratio of 4 has the greater head loss.
Problem 8.61
A viscous oil with a specific gravity SG = 0.85 and a viscosity of 0.10 Pa ⋅ s flows from tank
A to tank B through the six rectangular slots indicated in the figure below. If the total
mm3
flowrate is 30
and minor losses are negligible, determine the pressure in tank A .
s
Same elevation
A
B
Each slot
3 mm
× 1 mm
0.6 m
a
a
Section
a−a
Solution 8.61
3
3
1  mm 3   1m 
−9 m
For each slot, Q =  30
so that
=
×
5
10

6 
s   103 mm 
s
m3
Q
m
s
= 1.67 × 10 −3
V= =
2
A
s
 1m 
3mm 2  3

 10 mm 
Thus, with
5 ×10 −9
(
)
(
)
2
kg 
kg
4A 4 3 mm

ρ = SG ρH2O = 0.85 1000 3  = 850 3 and Dh =
=
= 1.5 mm
P (6 + 2) mm
m 
m

we obtain,
 1m 
kg  

−3 m 
 850 3  1.67 × 10
 (1.5 mm ) 

s 
ρVDh 
m 
 1000 mm  = 0.0213 < 2100
=
Re h =
N ⋅s
μ
0.1 2
m
Thus, the flow is laminar with f Re h = 69.3 (see Table 8.3 Friction Factors for Laminar
a 1
Flow in Noncircular Ducts [Data from Ref. 18] with = ).
b 3
a
a
= 0.50 as
Interpolate between = 0.25 and
b
b
69.3
(0.333 − 0.25)
f Re h = 72.9 −
= 3250
( 72.9 − 62.2 ) = 69.3 or f =
(0.50 − 0.25)
0.0213
pA VA2
pB VB 2
V2
+
+ zA =
+
+ zB + f
The energy equation
γ
2g
γ
2g
Dh 2 g
with zA = zB , pB = 0 ,VA = VB = 0 gives
1
0.6 m  1  
kg  
−3 m 
pA = f
ρV 2 = (3250)
   850 3  1.67 × 10

0.0015 m  2  
s 
Dh 2
m 
N
= 1540 2 = 1.54 kPa
m
2
Problem 8.62
Air at standard temperature and pressure flows at a rate of 7.0 cfs through a horizontal,
galvanized iron duct that has a rectangular cross-sectional shape of 12 in. by 6 in.
Estimate the pressure drop per 200 ft of duct.
Solution 8.62
For a horizontal duct, Δp = γ hL = f
Q
1
ρV 2 where V =
Dh 2
A
or
ft3
7
VDh
ft
s
= 14.0
V=
and Reh =
ν
s
 1ft 2 
(12 in.)( 6 in.) 

2
 144 in. 
with
(
)
2
4A 4 0.5 ft
Dh =
=
= 0.667 ft
P
(2 + 1) ft
Thus,
ft
14.0 ( 0.667 ft )
s
= 5.95 × 10 4
Re h =
2
ft
1.57 × 10 −4
s
Also, for galvanized iron, ε = 0.0005 ft, or
ε
Dh
From the figure below, we obtain f = 0.0227
=
0.0005ft
= 0.000750
0.667ft
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
6
104
2(105)
8
4
6
105
2(106)
8
106
4
6
2(107)
8
107
ρ VD
Re = _____
μ
Thus, from Eq. (1), with
Δp = ( 0.0227 )
= 200 ft ,
2
200 ft 1 
slug 
ft 
lb
2.38 ×10 −3 3  14.0  = 1.59 2 = 0.0110 psi

0.667 ft 2 
s
ft  
ft
4
6
8
0.00001
ε
__
D
Problem 8.63
Water at 20 °C flows through a concentric annulus of inner diameter D1 = 2.0 cm and
outer diameter D2 = 4.0 cm. The surface roughness is 0.002 cm. Calculate the frictional
pressure loss per unit length for a flowrate of 0.01 m3 /s.
Solution 8.63
The frictional pressure loss is
V2
ΔpL = ρ ghL = ρ f
Dh 2
where V =
(
4Q
π D22 − D12
)
and Dh is the hydraulic diameter,
or Dh =
Δ pL =
4A
= D2 − D1, so
P
8ρ f Q 2
(
π 2 ( D2 − D1 ) D22 − D12
)
2
The Reynolds number is found using

m3  
cm 
4  0.01
  100


s 
s 
VDh 4Q ( D2 − D1 )
4Q

Re h =
=
=
=
ν
πν ( D2 + D1 )

m2 
πν D22 − D12
π 1.0 × 10−6
 ( 4.0 + 2.0 ) cm
s 

(
)
= 2.12 ×105.
Now
ε
Dh
=
ε
D2 − D1
=
0.002 cm
= 0.001.
( 4.0 − 2.0 ) cm
Assuming that the friction factor for turbulent flow through an annulus is the same as that
for a circular pipe with the same hydraulic diameter, the Moody chart gives f = 0.021, so
ΔpL
=
8 ρ fQ 2
(
π 2 ( D2 − D1 ) D22 − D12
)
2
5

kg 
m3  
cm   N ⋅ s 2 

8  998 3  ( 0.021)  0.01
100
 

 

s  
m   kg ⋅ m 
m 


=
2
N 

π 2 ( 4.0 − 2.0 ) cm 4.02 − 2.02 cm 4 103 2

m kPa 

ΔpL
kPa
.
= 59.0
m
(
)
Problem 8.64
Air at standard conditions flows through a horizontal 1 ft by 1.5 ft rectangular wooden
ft3
duct at a rate of 5000
. Determine the head loss, pressure drop, and power supplied by
min
the fan to overcome the flow resistance in 500 ft of the duct.
Solution 8.64

ft3   1min 
5000



min   60 s 
V2
Q 
ft
hL = f
= 55.6
, where V = =
Dh 2 g
s
A
(1ft )(1.5 ft )
and Dh =
4A 4 (1ft ) (1.5ft)
=
= 1.2 ft
P 2 [1ft + 1.5 ft ]
Also,
ft 

55.6  (1.2 ft )

VDh 
s
Re h =
=
= 4.25 ×105 and from the table below
2
ν
ft
1.57 ×10−4
s
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
ε ≈ 0.0006 ft to 0.003 ft . Use an “average” ε = 0.0018 ft so that
ε
0.0018ft
=
= 0.0015 . Thus, from the figure below, f = 0.022
Dh
1.2 ft
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
104
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
2
ft 

55.6 

 500 ft  
s
= 440 ft
or hL = (0.022) 

 1.2 ft  2  32.2 ft 


s2 

p1 V12
p2 V22
+
+z =
+
+ z2 + hL , where z1 = z2 and V1 = V2 .
For this horizontal pipe,
2g
γ 2g 1 γ
lb 
lb

Thus, p1 − p2 = γ hL =  7.65 ×10 −2 3  ( 440 ft ) = 33.7 2 = 0.234 psi
ft 
ft





ft 3   1min  
lb  
ft ⋅ lb  
1hp


P = γ QhL = Q ( p1 − p2 ) =  5000
33.7
2810
=



 

ft ⋅ lb  
min   60 s  
s  
ft 2  

550


 
s  
or
P = 5.11 hp
ε
__
D
Problem 8.65
Assume a car’s exhaust system can be approximated as 14ft of 0.125-ft- diameter cast-iron
pipe with the equivalent of six 90 flanged elbows and a muffler. The muffler acts as a
resistor with a loss coefficient of K L = 8.5 . Determine the pressure at the beginning of the
exhaust system if the flowrate is 0.10 cfs , the temperature is 250 F , and the exhaust has
the same properties as air.
Solution 8.65
MIDAS
KLm = 8.5
V2
(2)
(1)
KLel = 0.3
KL el = 0.3 (We obtain from Table B.2 Loss Coefficients for Pipe Components
(hL = K L
V2
) [Data from Refs. 5, 10, 27])
2g
2
p1 V12
p V2

V
+
+ z1 = 2 + 2 + z2 +  f +  K L  2 , where z1 = z2 , p2 = 0 ,
γ 2g
γ
2g
 D
 2g
and
ft3
0.1
Q
ft
s
V = V1 = V2 = =
= 8.15
A π 0.125 ft 2
s
(
)
4
Thus,
lb  
in.2 

14.7
144



in.2  
ft 2 

slug
p

1 2
p1 =  f +  KL  ρV , where ρ =
=
= 1.74 ×10−3 3
RT 
ft
 D
2
ft ⋅ lb 
460 + 250 ) ° R
1716
° (
slug ⋅ R 

ε 0.00085 ft
=
= 0.0068 (see the table below)
Also,
D
0.125ft
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
slug 
ft 

1.74 ×10 −3 3   8.15  ( 0.125 ft )

ρVD 
s
ft  
=
= 3770
so that with Re =
μ
−7 lb ⋅ s
4.7 ×10
ft 2
We obtain from the figure below, f = 0.047
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
4
6
106
107
ρ VD
Re = _____
μ
Hence,

 1 
ft 
 14 ft 
−3 slug  
p1 =  0.047 
 + 6(0.3) + 8.5     1.74 × 10
 8.15 

3
s
ft  
 0.125ft 

 2 
lb
= 0.899 2
ft
2(107)
8
2
4
6
8
0.00001
ε
__
D
Problem 8.66
The pressure at section (2) shown in the figure below is not to fall below 60 psi when the
flowrate from the tank varies from 0 to 1.0 cfs and the branch line is shut off. Determine
the minimum height, h , of the water tank under the assumption that (a) minor losses are
negligible, (b) minor losses are not negligible
10 ft
h
6 ft
All pipe is 6-in.-diameter plastic
( ε /D = 0), flanged fittings
Branch line
(2) Main
line
600 ft
with 15
90° elbows
900 ft
Solution 8.66
With section (1) at the water tank surface,
2
p1 V12
p V2

V
+
+ z1 = 2 + 2 + z2 +  f +  K L  2 , where p1 = 0, V1 = 0, z1 = 16ft + h
γ 2g
γ
2g
 D
 2g
and z2 = 0 Thus, with V = V2
16 + h =
2
p2 V 2 
V
. Note: h mast be no less than that with p2min = 60 psi
+
+  f +  KL 
γ 2g  D
 2g
ft3
Q
ft
s
=
= 5.09
and Qmax = 1cfs , or V2 = V =
2
A2 π  6 
s
ft


4  12 
1
Hence,
2
lb  
in.2  

ft 




60
144




  5.09 s 
 h + 6 + 600 + 900 
ft 2  
 in.2  
h = −16ft +
+ 1 + f 
 +  KL 
lb
6

 2  32.2 ft 


62.4 3


12



 
ft
s2 
or


 1506 + h 
h = 122.5 + 1 + f 
 +  K L  ( 0.402 ) ft , where h ∼ ft
 0.5 


ft  6 

5.09  ft 

ε
VD 
s  12 
= 0 and Re =
=
= 2.10 ×105 , we obtain
With
2
ν
D
ft
1.21×10−5
s
f = 0.0155 (see the figure below)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
103
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
(a) Neglect minor losses (  K L = 0 ) :
From Eq. (1),

 1506 + h  
h = 122.5 + 1 + ( 0.0155 ) 
  ( 0.402 )
 0.5  

or h =143 ft
(b) Include minor losses:
 KL = KL
entrance
+ 15K Lelbow + K Ltee = 0.5 + 15(0.3) + 0.2 = 5.2 (see Table B.2 Loss
V2
Coefficients for Pipe Components (hL = K L
) [Data from Refs. 5, 10, 27], assume
2g
flanged fillings)
ε
__
D
Thus, from Eq. (1),


 1506 + h 
h = 122.5 + 1 + ( 0.0155 ) 
 + 5.2  ( 0.402 )
 0.5 


or
h =146 ft
Note: For this case, minor losses are not very important.
Problem 8.67
The pressure at section (2) shown in the figure below is not to fall below 60 psi when the
flowrate from the tank varies from 0 to 1.0 cfs and the branch line is open so that half of
the flow from the tank goes into the branch, and half continues in the main line. Determine
the minimum height, h , of the water tank under the assumption that (a) minor losses are
negligible, (b) minor losses are not negligible
(1)
•
10 ft
h
6 ft
(a)
All pipe is 6-in.-diameter plastic
(ε /D = 0), flanged fittings
Branch line
(b)
(2) Main
line
600 ft
with 15
90° elbows
900 ft
Solution 8.67
For the flow from (1) to (2):

V 2 
V 2
p1 V12
p V2
+
+ z1 = 2 + 2 + z2 +  fa a +  K La  a +  fb b +  K Lb  b
γ 2g
γ
2g
 Da
 2 g  Db
 2g
(1)
where ( )a and ( )b denote pipes “a” and “b” as indicated in the figure.
Thus, with p1 = 0, V1 = 0, z1 = 16ft + h, z2 = 0 , and p2 = 60 psi . Also,
ft3
ft3
0.5
Q
Q
ft
ft
s
s
Va = a =
= 5.09 , Vb = b =
= 2.55 , Eq. (1) becomes
2
2
Aa π  6 
Ab π  6 
s
s
 ft 
 ft 
4  12 
4  12 
1
2
lb  
in.2  

  5.09 ft 


 60 2   144 2  

 
 h + 6 + 600 
ft 
 in.  
s
16 + h =
K
+  1 + fa 
+
  La 
lb
6

 2  32.2 ft 


62.4 3

12



 
ft
s2 
2
ft 
 


  900 
  2.55 s 
+  fb 
 +  K Lb 
6
 
 2  32.2 ft 


  12 
 
s2 
or


 606 + h 
h = 122.5 + 1 + fa 
 +  K La  (0.402)+ 1800fb + K Lb (0.101) , where h ∼ ft
 0.5 


(
)
(2)
ft  6 

5.09  ft 

VD
ε
s  12 
= 2.10 ×105 , and
= 0 , Rea = a a = 
With
2
ν
D
ft
1.21×10−5
s
Vb Db 1
= Rea = 1.05 ×105 , we obtain fa = 0.0155 and fb = 0.0175 (see the figure
Reb =
2
ν
below)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
103
6
2(104)
8
4
6
104
2(105)
8
4
105
6
2(106)
8
4
106
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
(a) Neglect minor losses
(  KL =  KL = 0) :
a
b
From Eq. (2),

 606 + h  
h = 122.5 + 1 + (0.0155) 
  (0.402)+ (1800 (0.0175) ) (0.101)
 0.5  

or
h =135 ft
(b) Include minor losses:
 KL = KL
a
entrance
+ 15 K Lelbow = 0.5 + 15(0.3) = 5.0 (see Table B.2 Loss Coefficients for Pipe
V2
Components ( h L = K L
) [Data from Refs. 5, 10, 27], assume flanged fillings)
2g
ε
__
D
and
 K L = K L = 0.2
b
tee
From Eq. (2),


 606 + h 
h = 122.5 + 1 + (0.0155) 
 + 5.0  (0.402)+ (1800 (0.0175)+0.2 ) (0.101)
 0.5 


or
h =137 ft
Note: For this case, minor losses are not very important.
Problem 8.68
1
The -in.- diameter hose shown in the figure below can withstand a maximum pressure of
2
200 psi without rupturing. Determine the maximum length, , allowed if the friction
factor is 0.022 and the flowrate is 0.010 cfs. Neglect minor losses.
Nozzle tip diameter = 0.30 in.
Q=
0.010 cfs
D = 0.50 in.
10 ft
Water
𝓵
L
3 ft
Pump
Solution 8.68
Nozzle tip diameter = 0.30 in.
Q=
0.010 cfs
•
(2)
D = 0.50 in.
10 ft
Water
L
(1)
•
3 ft
𝓵
Pump
p1 V12
p V2
V2
,
+
+ z1 = 2 + 2 + z2 + f
D 2g
γ 2g
γ
2g
where z1 = 0, z2 = 10 ft, p1 = 200 psi
ft3
ft3
0.01
Q
ft
Q
ft
s
s
=
= 20.4
=
= 7.33 , and V2 =
p2 = 0, V1 =
2
A2 π  0.3 
s
A1 π  0.5 2 
s
ft 

ft


4  12

4  12 
0.01
Thus, with f = 0.022 , Eq. (1) becomes (using V = V1 )
2
2
2
lb  
in.2  

ft 
ft 
ft 




200
144


7.33 
7.33 


 20.4 

 
in.2  
ft 2  

s
s
s

+
=
+ 10 ft + 0.022 

lb
0.5
ft 
ft 
ft



62.4 3
ft  2  32.2 
2  32.2 2  2  32.2 2 
 12  
ft
s 
s 
s2 


or
= 1012 ft
(1)
Problem 8.69
The hose shown in the figure below will collapse if the pressure within it is lower than
10 psi below atmospheric pressure. Determine the maximum length, , allowed if the
friction factor is 0.015 and the flowrate is 0.010 cfs . Neglect minor losses.
Nozzle tip diameter = 0.30 in.
Q=
0.010 cfs
D = 0.50 in.
Water
(1)
•
10 ft
𝓵
L
(2)
•
3 ft
Pump
Solution 8.69
p1 V12
p2 V2 2
LV2
+
+z =
+
+ z2 + f
,
γ 2g 1 γ
2g
D 2g
where p1 = 0, V1 = 0, z1 = 3ft , z2 = 0, p2 = −10
(1)
lb
in.2
, and
ft3
Q
ft
s
V2 =
=
= 7.33 = V
A2 π  0.5 2 
s
ft 

4  12

0.01
Thus, with f = 0.015 , Eq. (1) becomes
2
lb  
in.2  


   7.33 ft 
10
144
−





 L   
in.2  
ft 2  

s
3ft =
+ 1 + ( 0.015 ) 

lb
0.5   
ft


62.4 3
ft   2  32.2 
 12   

ft
s2 
or
L = 84.0 ft
Problem 8.70
According to fire regulations in a town, the pressure drop in a commercial steel horizontal
gal
. If the water
pipe must not exceed 1.0 psi per 150 ft of pipe for flowrates up to 500
min
temperature is above 50 F , can a 6-in.- diameter pipe be used?
Solution 8.70
Determine the pressure drop in a 6-in.- diameter pipe.
p1 V12
p V2
V2
+
+ z1 = 2 + 2 + z2 + f
, where V1 = V2 and z1 = z2 .
γ 2g
γ
2g
D 2g
Thus
p1 − p2
γ
= f
ε 
V2

, where f = f  Re,  .
D
D 2g

From the table below, ε = 0.00015 ft so that
(1)
ε
D
=
1.5 × 10 −4
= 3 × 10 −4
 6 
 ft 
 12 
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.00015
0.000005
0.0 (smooth)
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.045
0.0015
0.0 (smooth)
The largest p1 − p2 will occur with the largest f , which occurs with
the smallest Re , or largest ν . Since the viscosity of water increases as
the temperature decreases, we consider the coldest case T = 50 F .
From Table B.1 Physical Properties of Water (BG/EE Units), at
ft 2
lb
50 F, γ = 62.41 3 and ν = 1.407 ×10 −5
s
ft
Also, V =
Q
=
A
gal   1min  
in.3   1ft 3 

500
231






min   60 s  
gal   1728 in.3 

π6

 ft 
4  12 
2
= 5.67
ft
s
f
Re
Thus,
ft  6 

5.67  ft 

VD 
s  12 
Re =
=
= 2.01×105
2
ν
ft
1.407 ×10−5
s
ε
Hence, with Re = 2.01×105 and
= 3 ×10−4 , we obtain from the figure below, f = 0.018
D
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
f
0.004
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
104
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Therefore, from Eq. (1),
2
ft 

5.67 

150ft
(
)
p1 − p2
s
= 0.018
= 2.70 ft
ft 
γ
6  
 ft  2  32.2 2 
 12  
s 
so that
lb 
lb  1ft 2 

p1 − p2 = ( 2.70 ft )  62.4 3  = 168 3 
 = 1.17 psi > 1.0 psi
ft 
ft  144 in.2 

A 6-in.- diameter pipe requires slightly more than the allowed 1.0 psi per 150 ft . No, a 6-in.
pipe cannot be used. The minimum diameter can be shown to be D = 0.513ft = 6.37 in.
Problem 8.71
As shown in the figure below, water “bubbles up” 3 in. Above the exit of the vertical pipe
attached to three horizontal pipe segments. The total length of the 0.75-in.-diameter
galvanized iron pipe between point (1) and the exit is 21 in. Determine the pressure needed
at point (1) to produce this flow.
(1)
3 in.
4 in.
Solution 8.71
V
KL = 1.5
3 in.
• (2)
4 in.
• (3)
(1)
p1 V12
p V2
+
+ z1 − hL = 2 + 2 + z2
γ 2g
γ 2g
where z1 = 0, p2 = 0, V2 = 0. Thus,
p1
γ
= z2 + hL −
V12
where V1 = V3 = V
2g
(1)
With no head loss from (3) to (2) and p2 = p3 = V2 = 0 we obtain
V32
ft  3 
ft

+ z3 = z2 , or V3 = 2 g ( z2 − z3 ) = 2  32.2 2   ft  = 4.01
2g
s
s   12 

Thus,
ft  0.75 
4.01 
ft 
VD V3D
s  12
 = 2.07 ×10 4
Re =
=
=
2
v
v
ft
1.21×10 −5
s
and
ε
D
=
0.0005 ft
= 0.008 (see the table below), so that (see the figure below) f = 0.039
 0.75 
ft 

 12

Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7) and Colebrook (Ref. 8)]
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel
or wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.00015
0.000005
0.0 (smooth)
0.045
0.0015
0.0 (smooth)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
ρ VD
Re = _____
μ
Also, hL = f
 V2
V2
where  K L = 3(1.5) = 4.5
+  KL
D 2g
2g
4
6
2(107)
8
107
4
6
8
0.00001
ε
__
D
Hence, Eq. (1) becomes
2
V12
 
V
= z2 +  f +  K L 
−
γ
 b
 2g 2g
p1
where V1 = V
or
2
ft 

4.01 



p1 7
21 in.
s
= ft + 0.039
+ 4.5 − 1 
= (0.583 + 1.147)ft = 1.73 ft
γ 12
0.75 in.

 2  32.2 ft 


s2 

Thus,
lb 
lb

p1 =  62.4 3  (1.73 ft) = 108 2 = 0.750 psi
ft 
ft

Problem 8.72
m3
,
s
what is the maximum length inlet pipe, l, that can be used without cavitation occurring?
Water at 10°C is pumped from a lake as shown in the figure. If the flowrate is 0.011
Length 𝓵
D = 0.07 m
ε = 0.08 mm
Elevation
650 m
Elevation
653 m
Q=
0.011 m3/s
Solution 8.72
Length 𝓵
Elevation
650 m
(1)
•
D = 0.07 m
ε = 0.08 mm
(2)
•
Elevation
653 m
Q=
0.011 m3/s
2
p1 V12
p2 V22
 
V
+
+z =
+
+ z2 +  f +  K L 
, where p1 = 101 kPa, z1 = 650 m
γ 2g 1 γ 2g
 b
 2g
(1)
V1 = 0, V2 = V, z2 = 653 m, and from Table B.2 Physical Properties of Water (SI Units)
p2 = pv = 1.228 kPa
m3
0.011
Q
s = 2.86 m so that
Also, V = =
s
A π (0.07 m)2
4
m

286
 (0.07 m)
VD 
s 
Re =
=
= 1.53 ×105. With this Re and from Table 8.1 Equivalent
2
v
m
1.307 ×10 −6
s
Roughness for New Pipes [Adapted from Moody (Ref. 7) and Colebrook (Ref. 8)]
with
ε
D
=
0.08 mm
= 0.00114 , we obtain f = 0.0216 (see the figure below)
70 mm
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
103
6
2(104)
8
104
4
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
107
ρ VD
Re = _____
μ
Hence, with  K L = 0.8 for the entrance, Eq. (1) becomes
2
m
N

2.86 
(101 − 1.228) ×10

3


s 
m + 650 m = 653 m + 1 + (0.0216)    + 0.8 




N
 0.07 m 

 2  9.81 m 
9.80 × 103 2


m
s2 

3
or
ℓ = 50.0 m
6
8
0.00001
ε
__
D
Problem 8.73
At a ski resort, water at 40  F is pumped through a 3-in.-diameter, 2000-ft -long steel pipe
from a pond at an elevation of 4286 ft to a snow-making machine at an elevation of 4623 ft
ft 3
at a rate of 0.26
. If it is necessary to maintain a pressure of 180 psi at the snow-making
s
machine, determine the horsepower added to the water by the pump. Neglect minor losses.
Solution 8.73
pump
(2)
(1)
p1 V12
p V2
 V2
1b
, where p2 = 180 2 , p1 = 0, V1 = 0 (1)
+
+ z1 + hp = 2 + 2 + z2 + f
D 2g
γ 2g
γ 2g
in
ft3
Q
s = 5.30 ft . From Table 8.1 below, ε = 0.00015 ft = 6 ×10 −4
and V = V2 =
=
3
π 2 π  3 2
D
s
ft
D
ft


12
4
4  12 
0.26
Equivalent Roughness for New Pipes [Adapted from
Moody (Ref. 7) and Colebrook (Ref. 8)]
Equivalent Roughness, ε
Pipe
Feet
Millimeters
Riveted steel
Concrete
Wood stave
Cast iron
Galvanized iron
Commercial steel or
wrought iron
Drawn tubing
Plastic, glass
0.003–0.03
0.001–0.01
0.0006–0.003
0.00085
0.0005
0.9–9.0
0.3–3.0
0.18–0.9
0.26
0.15
0.00015
0.000005
0.0 (smooth)
0.045
0.0015
0.0 (smooth)
ft  3 

5.30  ft 

VD 
s  12 
=
= 7.96 ×10 4 so that f = 0.0212 (see the figure below)
Also, Re =
2
v
ft
1.664 ×10 −5
s
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
103
6
2(104)
8
4
104
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
107
ρ VD
Re = _____
μ
Thus, from Eq. (1)
hP =
2
 V

+ z2 − z1 + 1 + f 
γ
D  2g

p2
or
2
1b  
1b 



   5.30 ft 
 180 2   144 2 



2000 ft   
s
in.
in.




hP =
+ 4623 ft − 4286 ft + 1 + (0.0212) 

lb
3
ft


62.4 3
ft   2  32.2 
 12
 

ft
s2 
Hence, hP = 827 ft so that


3


1b 
ft
ft·1b 
1 hp 


(827 ft) =  13420
P = γ QhP =  62.4 3   0.26


 = 24.4 hp

s 
s   550 ft·lb 

ft  

s 

6
8
0.00001
ε
__
D
Problem 8.74
Crude oil having a specific gravity of 0.80 and a viscosity of 6.0 × 10−5 ft2/s flows through a
pumping station at a rate of 10000 barrels/hr. The oil then flows through 120000 ft of 24-in.
ID, horizontal, commercial steel pipe, enters another pumping station at 20 psig, and leaves
at a discharge pressure equal to the discharge pressure of the previous pumping station.
Calculate the discharge pressures and the power supplied to the crude oil at the second
pumping station. See the figure below.
L = 120000 ft
4
3
2
1
I.D. = 24 in. = 2 ft
·
Wp
·
Wp
Solution 8.74
Assume constant oil density and apply the mechanical energy equation from point 2 to
point 4.
p2
ρ
+
V22
p V2
+ gz2 − ws = 4 + 4 + gz4 + ghL .
2
ρ
2
Now p2 = p4, the continuity equation gives V2 = V4, and z2 = z4 for a horizontal pipeline.
Then
ws = −ghL or wp = ghL
Since ws = −wp (from W s = −W p). Now
L V2 f L Q 
w p = ghL = f
=
 
D 2
2 D A
2
and
ρ fLQ3 8ρ fLQ3
 p = ρQw p =
.
W p = mw
=
π 2 D5
2DA2
The friction factor is found using Table 8.1 to give
ε
D
=
0.00015 ft
= 0.000075
2 ft
There are 42 U.S. gallons in one barrel. The Reynolds number is
Re =
VD 4Q
=
=
v
π Dv
= 1.65 ×105
barrels  
gal   ft3   hr 

4 10000
42




hr   barrel   7.48 gal   3600 s 

(
π (2 ft) 6.0 ×10 −5 ft 2 / s
)
The Moody chart gives f = 0.016. Then
W p =

lbm 
barrels 
gal   ft3  

8(0.8)  62.4 3  (0.016)(120000 ft) 10000

 42

hr  barrel   7.48gal  
ft 


π
2
3
3
5  32.2 ft ⋅ lbm  3600 s   550 ft ⋅ lb 
(2 ft) 
  hr   s ⋅ hp 
2


lb ⋅ s
 

or
W = 520hp.
p
The pressure at point 2 (or 4) is found by applying the mechanical energy equation between
points 2 and 3.
p 2 V22
p V2
+
+ gz2 − ws = 3 + 3 + gz3 + ghL .
ρ
ρ
2
2
Now ws = 0, z2 = z3 and the continuity equation gives V2 = V3. Then
 LV2 
ρ f LQ 2
8ρ f LQ 2
p2 = p3 + ρ ghL = p3 + ρ  f
p
p
=
+
=
+
3
3
 D 2 
2DA2
π 2 D5


The numerical values give
p2 = 20psig +

lbm 
barrels  
gal   ft3  

8(0.8)  62.4 3  (0.016)(120000 ft)  10000

  42

hr   barrel   7.48gal  
ft 


π
p2 = p4 = 147 psig.
2
 144 in2 


  ft 2 
 32.2 ft ⋅ lbm   3600 s 
(2 ft)5 
  hr 
2

lb ⋅ s

2
2
Problem 8.75
A motor-driven centrifugal pump delivers 15°C water at the rate of 10 m3/min from a
reservoir, through a 2500-m-long, 30-cm I.D. plastic pipe, to a second reservoir. The water
level in the second reservoir is 40 m above the water level in the first reservoir. The pump
efficiency is 75%. Find the motor output power. The pipe entrance is square edged.
Solution 8.75
2
•
40 m
1
•
00
m
25
Assume constant density. Write the mechanical energy equation from (1) to (2) with ht = 0
to get
p1 V12
p 2 V22
+
+ g z1 + ghp =
+
+ g z2 + ghL
ρ
2
ρ
2
(1)
Now V1 ≈ V2 ≈ 0 and use p1 = p2 = patm = 0. Then
ghp = g (z2 – z1) + ghL
where
2
 
V
ghL =  f + K Lent + K Lexit 
 D
 2
so
2
 
V
ghp = g ( z2 − z1 ) +  f + K Lent + K Lexit 
 D
 2
For a square-edged entrance, Fig. 8.22 and Section 8.4.2 give KLent = 0.50 and KLexit = 1.0.
The velocity is
 m3   min 
 10


min   60 s 
m
Q

V= =
= 2.36 .
A
s

m2
 π 
(30 cm)2   

2
 4   10000 cm 
For 15°C water, interpolating Table B.2 gives
 m 
m

 2.36  (30 cm) 

s 
100 cm 
VD 

=
= 6.15 × 105
Re =
2
v

−6 m 
1.15 ×10

s 

The Moody chart gives f = 0.013. The energy equation is
2
m

2.36 



m
(2500 m)
m2

s 
ghp =  9.81 2  (40 m) +  0.013
+ 0.5 + 1.0  
= 698 2 .
(0.30 m)
2
s 
s



The motor output power is
kg   m3   min  
m2 

999
10
698






3 
2 

min   60 s  
ρ
Qghp 
m
s




Woutput =
=
0.75
η pump
2
kg ⋅ m
N⋅m
,
W output = 1.55 ×105
= 1.55 ×105
3
s
s
and
W output = 155 kW
Problem 8.76
An emergency flooding system for a nuclear reactor core is shown in the figure below. Find
the power input required to flood the core at the rate of 5000 gal/min. Assume a squareedged entrance, 60°F water, and threaded connections.
patm
50000
gallons
water
750 ft of 3-in. schedule 160 stainless
steel pipe (actual I.D. is 2.625 in.), three
gate valves, seven 90° regular elbows
2000
psig
Core
Solution 8.76
Assume constant density. Write the mechanical energy
equation from the reservoir level 1 to the pipe discharge
2 with ht = 0 to get
p1 V12
p V2
+
+ g z1 + ghp = 2 + 2 + gz2 + ghL
ρ
2
ρ
2
Now V1 ≈ 0, assume z1 – z2 ≈ 0, and choose p1 = patm = 0. The
energy equation is then
ghp =
p2
ρ
+
V22
+ ghL
2
patm
• 1
50000
gallons
water
750 ft of 3-in. schedule 160
stainless steel pipe, three gate
valves, seven 90° standard
elbows
• 2000
2
psig
where
2
 
V
ghL =  f + 3K Lgate + K Lelb + K Lent  2
 D
 2
so
ghp =
2


V
+  1 + f + 3K Lgate + 7 K Lelb + K Lent  2
D
ρ 
 2
p2
For threaded connections and a square-edged inlet, Table 8.2 and Fig. 8.22 give
KLgate = 0.15, KLelb = 1.5, KLent = 0.5.
Core
Tables 8.1 and B.1 for 60 °F water give
ε
D
=
0.00015 ft
= 0.000686
 2.625 
ft 

 12



ft3


gal 
ft3

s


11.14
,
=
Q =  5000

min   448.8 gal 
s


min 


ft3 
11.14


s 
Q
ft

=
V= =
296
,
A π  2.625 2
s
ft 

4  12

and
ft  2.625 

296 
ft 

VD 
s  12
 = 5.35 ×106.
Re =
=
2
V

−5 ft 
1.21×10

s 

The Moody chart gives f = 0.018. The energy equation gives
lb  
in2 

2000
144





in2  
ft 2 
ghp =
lbm 

 62.4 3 
ft 



ft 
296 
2




(750 ft)
s   lb ⋅ s
+ 1 + (0.018)
+ 3(0.15) + 7(1.5) + 0.5  


2
32.2 ft ⋅ lbm 
 2.625 



ft 



 12

ft ⋅ lb
= (4615 + 100900)
lbm
ft ⋅ lb
= 105515
lbm
The power input (to the water) is
lbm  
ft3  
ft ⋅ lb 

−  62.4 3   11.14
  97855


s 
lbm 
ft  


W = − ρQghp =

ft ⋅ lb 
 550 s ⋅ hp 


or
W = −133360 hp
Problem 8.77
A hydraulic turbine takes water from a lake with the piping system shown in the figure
below. Find the head ht available to the turbine for flowrates of 0, 5000, 10000, 15000, and
20000 ft3/min.
1100 ft
Equivalent length of
900 ft of 5-ft inside
diameter concrete pipe
I.D. = 5 ft
Hydraulic turbine
Solution 8.77
Assume constant fluid density and steady state. Apply the mechanical energy equation
from the lake’s water surface (1) to the turbine outlet (2).
p1 V12
p2 V22
+
+ gz1 + ghp =
+
+ gz2 + ght + ghL .
ρ
2
ρ
2
1
•
1100 ft
Equivalent length of
900 ft of 5-ft inside
diameter concrete pipe
I.D. = 5 ft
Hydraulic turbine
Now p1 = p2 = patm, V1 ≈ 0, and hp = 0 to get
ht = ( z1 − z2 ) −
Using V2 =
V22
V2
 V22
− hL = ( z1 − z2 ) − 2 − f
.
D 2g
2g
2g
4Q
π D2
,
8Q 2 

ht = ( z1 − z2 ) − 2 4 1 + f  .
D
π D g
•
2
The known numerical values give
2
 ft3   min 2
8Q
 min   60 s  
(900 ft) 

1+ f
ht = 1100 ft − 


ft  
(5 ft) 
2
4
π (5 ft)  32.2 2 
s 

= 1100 ft − 1.119 ×10−8Q 2 (1 + 180 f )ft
Assuming 60 °F water, noting Q is in
ft 3
, and using property tables,
min
 ft3   min 
4 Q
 min   60 s 
VD
4Q


=
=
= 351 Q
Re =
2
π Dν
v

−5 ft 
π (5 ft)  1.21×10

s 

and
ε
D
=
0.005 ft
= 0.001.
5 ft
Using the Moody chart, we tabulate the available turbine head ht versus Q.
Q
(ft /min)
Re
f
Available head,
ht (ft)
0
0
0.0198
1100
5000
1.755×10
0.0197
1099
10000
3.51×10
0.0197
1095
15000
5.265×10
0.0197
1089
20000
7.02×10
0.0197
1080
3
Problem 8.78
ft
as shown in the figure
s
below. The relative roughness of the pipe is 0.004, and the loss coefficient for the exit is 1.0.
Determine the height, h, to which the water rises in the piezometer tube.
Water flows through a 2-in.-diameter pipe with a velocity of 15
Open
h
8 ft
2 in.
15 ft/s
8 ft
Solution 8.78
Open
(2)
•
h
8 ft
2 in.
15 ft/s
V
•
(1)
8 ft
V12
p
V2
p
− hL = 2 + z2 + 2 , where 1 = h , z1 = 0, p2 = 0, z2 = 8 ft, V 2 = 0 and
γ
2g
γ
2g
γ
2
 
V
hL =  f + K L 
with V = V1 and KL = 1
 D
 2g
p1
+ z1 +
Thus
h+
2
V2  
V
−  f + KL 
= z2
2g  D
 2g
But Re =
ρVD
=
μ
(1)
slug  ft  2 
15  ft 
ft3  s  12  = 2.07 ×105
lb ⋅ s
2.34 ×10−5 2
ft
1.94
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
6
104
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Hence from the figure above with
2
ε
D
= 0.004, we obtain f = 0.029 so that Eq. (1) becomes

  ft 

  15 s 
8 ft
 = 8 ft
− 1 
h + 1 − 0.029
ft 
2   

 ft   2  32.2 2 

 12   
s 
or
h = 12.9 ft
ε
__
D
Problem 8.79
The figure below shows the 60°F water flowrates from the branches of a main supply line.
Find the total pressure drop (pA − pE) for soldered copper pipe. Assume that the loss
coefficient for each tee is based on the threaded type.
Main supply line, 4-in.
type L copper pipe
(actual I.D. is 3.905 in.)
A
Q
B
70 ft
100 gal/min
500 gal/min
C
140 ft
D
E
90 ft
150 gal/min
100 gal/min
Solution 8.79
Assume constant density and steady state. Applying the mechanical energy equation from
A to E gives
p A VA2
p E VE2
+
+ gzA + ws =
+
+ gzE + ghL AE
ρ
2
ρ
2
Now ws = 0 and assume zA = zE.
Then
 V 2 −VA2 
pA − pE = ρ  E
 + ρ ghL AE

2


where
2
2
2
 
 VAB  
 VBC  
 VCD
ghLAE =  f + K Ltee 
+  f + K Ltee 
+  f + K Ltee 
 D
AB 2
 D
BC 2
 D
 2
For soldered (similar to threaded) pipe, Table 8.2 gives KL tee = 0.90 for flow through the
line. For 60°F water,


ft3 / s
4(850 gal/ min) 
 449 gal/ min 
VD
4Q

 = 6.12 × 105 ,
=
=
Re AB =
2
π Dν
v
ft 
 3.905  
π
ft  1.21× l0 −5

s 
 12
 


ft3 / s
4(750 gal/ min) 
 449 gal/ min 
4Q

 = 5.40 ×105 ,
Re BC =
=
2
π Dν
ft 
 3.905  
ft  1.21× l0 −5
π

s 
 12
 
and


ft3 /s
4(600 gal/ min) 
 449 gal/min 
4Q

 = 4.32 ×105.
=
ReCD =
2
π Dν
ft 
 3.905  
ft  1.21× l0 −5
π


s 
 12

Table 8.1 gives
ε
D
=
(0.000005 ft)
= 0.000015
 3.905 
ft 

 12

So the Moody chart gives fAB = 0.0129, fBC = 0.0132, fCD = 0.0137.




 0.0129(70 ft)
 V 2  0.0132(140 ft)
V 2
AB
+ 0.90 
+
+ 0.90  BC
ghL AE = 
  3.905 ft 
 2
  3.905 ft 
 2



  12






  12


 0.0137 (90 ft)
V 2
+
+ 0.90  CD
  3.905 ft 
 2

  12


= 1.84VAB 2 + 3.29VBC 2 + 2.34VCD 2 .
Since VA = VAB, the mechanical energy equation gives
pA − pE = ρ (1.84 − 0.50)VAB 2 + 3.29VBC 2 + 2.34VCD 2 + 0.5VE 2 


The velocities are


ft3 / s
4(850 gal / min) 
 449 gal / min 
 4Q 
Q 

 = 22.8 ft / s,
VAB =   = 
=
2
2
A
 AB  π D AB
 3.905 
ft 
π
 12



ft3 / s
4(750 gal / min) 
 449 gal / min 
 4Q 

 = 20.1 ft / s,
=
VBC = 
2
2
 π D BC
 3.905 
ft 
π
 12



ft3 / s
4(600 gal / min) 
 449 gal / min 
 4Q 

 = 16.1 ft / s,
=
VCD = 
2
2
 π D CD
 3.905 
ft 
π
 12

and


ft3 / s
4(500 gal / min) 
 449 gal / min 
 4Q 

 = 13.4 ft / s
VE = 
=
2
2
 π D E
 3.905 
ft 
π
 12

For 60°F water,
2
lbm  

2
2
2
2  ft
pA − pE =  62.4 3  1.34 22.8 + 3.29 20.1 + 2.34 16.1 + 0.5 13.4
 s2
ft  

(

  ft 2 
lb ⋅ s2
∗
 32.2 ft ⋅ lbm   144 in2 



pA − pE = 36.6 psi
)
(
)
(
)
(
)
Problem 8.80
Water is pumped through a 60-m -long, 0.3-m -diameter pipe from a lower reservoir to a
higher reservoir whose surface is 10 m above the lower one. The sum of the minor loss
coefficients for the system is KL = 14.5. When the pump adds 40 kW to the water the
m3
flowrate is 0.20
. Determine the pipe roughness.
s
Solution 8.80
(2)
•
(1)
•
P
p1 V12
p2 V22
+
+z +h −h =
+
+ z2 , where p1 = p2 = 0, V1 = V2 = 0, z1 = 0, and z2 = 10 m
γ 2g 1 p L γ 2g
Thus, hp – hL = z2, where
N⋅m
s
=
= 20.4 m
hP =
3
γQ 

N
m

3

 9.80 ×10
  0.2
s 

m3  
40 × 103
W p
2
 
V
= 10 m
Hence, 20.4 m −  f +  K L 
 D
 2g
with V =
0.2
m3
s
Q
m
=
= 2.82
A π (0.3 m)2
s
4
Thus, from Eq. (1)
2
m

2.82 

  60 m 

s 
20.4 m −  f 
 + 14.5 
  0.3 m 
 2  9.81 m 


s2 

or
f = 0.0560
(1)
Also, Re =
ρVD
=
μ
999
kg 
m
2.82

 (0.3 m)
s 
m3 
= 7.55 ×105
N
s
⋅
1.12 ×10−3 2
m
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
104
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Thus, from the Moody chart (see the figure above) with Re = 7.55 × 105 and f = 0.0560, it
follows that
ε
D
= 0.028, or ε = 0.028(0.3 m) = 0.0084 m
f = 0.0560
ε = 0.028
D
Re = 7.55 × 105
ε
__
D
Problem 8.81
2

slugs
−5 ft 
and
v
5.2
10
=
×
Natural gas  ρ = 0.0044
 is pumped through a horizontal 6-in.3

s
ft


lb
diameter cast-iron pipe at a rate of 800
. If the pressure at section (1) is 50 psi (abs),
hr
determine the pressure at section (2) 8 mi downstream if the flow is assumed
incompressible. Is the incompressible assumption reasonable? Explain.
Solution 8.81
p1 V12
p V2
 V2
, where z1 = z2, and V1 = V2
+
+ z1 = 2 + 2 + z2 + f
γ 2g
γ 2g
D 2g
(1)
1b
0.222
lb  1 hr 
1b
ft3
s
=
0.222
,
or
Also, γ Q = 800 
Q
=
=
1.57

ft  
hr  3600 s 
s
s

−3 slug 
 32.2 2   4.4 × 10
2 
s 
ft 

ft3
Q
s = 8.00 ft
Thus, V = =
s
A π  6 2
 ft 
4  12 
1.57
ft   6 

8.00   ft 

ρVD VD 
s   12 
With Re =
=
=
= 7.69 × 10 4 and with
2
μ
v
ft
5.2 ×10 −5
s
ε
0.00085 ft
= 0.0017 (from Table B.1 Equivalent Roughness for New Pipes [Adapted
D
6 
 ft 
 12 
from Moody (Ref. 7) and Colebrook (Ref. 8)])
=
we obtain f = 0.0245. Thus, from Eq. (1)
ft 

(8 mi)  5280
2

1b
ft 
 1
mi  1 

−3 slug  
2
4.4 ×10
p2 = p1 − f
ρV = 50 2 − (0.0245)
  8.00 s 
0.5 ft
2 
D2
in.
ft3  

lb
= 50 psi − 291 2 = (50 − 2.02) psi
ft
or
p2 = 48.0 psi
Note:
p1 − p2 2.02 psi
=
= 0.0404, a 4.0% Charge in pressure.
50 psi
p1
Since p = ρRT, with T essentially constant, a small change in p gives a correspondingly
small change in ρ. Thus, the incompressible assumption is valid.
Problem 8.82
As shown in the figure below, a standard household water meter is incorporated into a lawn
irrigation system to measure the volume of water applied to the lawn. Note that these
meters measure volume, not volume flowrate. With an upstream pressure of p1 = 50 psi, the
meter registered that 120 ft3 of water was delivered to the lawn during an “on” cycle.
Estimate the upstream pressure, p1, needed if it is desired to have 150 ft3 delivered during an
“on” cycle. List any assumptions needed to arrive at your answer.
(1)
WATER
METER
Irrigation
system:
pipes, fittings,
nozzles, etc.
Solution 8.82
ℓ
fD
(1)
WATER
METER
Σ KL
Irrigation
(2)
system:
V2
•
pipes, fittings,
nozzles, etc.
The energy equation for this flow is
2
p1 V12
p V2
 
V
+
+ z1 −  f +  K L 
= 2 + 2 + z2
γ 2g
 D
 2g γ 2g
where z1 = z2, p2 = 0, V1 = V, and V2 =
(1)
A1
V1
A2
Thus, from Eq. (1)
2
 

 A1 
1
2
(2)
p1 = ρV1 f +  K L +   − 1
2
A2 
 D




V
, where V is the volume of water supplied during an “on” cycle and t is
But Q = AV
1 1=
t
the length of the cycle.
For a given system,  K L is independent of Q. Similarly, for large Re pipe flow, f is
2
 

 A1 
independent of Re (or Q). Thus,  f +  K L +   − 1 is constant, independent of Q.
 D

 A2 


Hence, from Eq. (2), if the length of the cycle is constant,
1

ρ1V12 
2
2
 V150   150 2
2
150 V1 )150 
=
=
 =
 =
 = 1.563
p1 )120 ft3 1
V1 )120 
V120   120 
2


ρ1V1 
2
120
p1 )150 ft3
or
p1 )150 = 1.563 p1 )120 = 1.563(50 psi) = 78.1 psi
Problem 8.83
A fan is to produce a constant air speed of 40 m/s throughout the pipe loop shown in the
figure below. The 3-m-diameter pipes are smooth, and each of the four 90° elbows has a
loss coefficient of 0.30. Determine the power that the fan adds to the air.
20 m
V = 40 m/s
D=3m
10 m
Fan
Solution 8.83
p1
γ
+ z1 +
V12
p
V2
− hL + hs = 2 + z2 + 2
2g
γ
2g
If we locate (1) and (2) at the same place, it follows that
p1 = p2, V1 = V2, and z1 = z2.
Thus,
where  KLi = 4(0.30) = 1.2
i
Also, Re =
ε
D
Fan
•
2
 
V
hs = hL =  f +  K Li 
 D
 2g
and
(1) ≡ (2)
ρVD
=
μ
kg 
m
40
 (3 m)
3
s 
m 
= 8.25 ×106
⋅
N
s
1.79 ×10−5
m2
1.23
= 0 so that from the figure below, f = 0.0083
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
4
105
6
2(106)
8
4
6
106
ρ VD
Re = _____
μ
Hence,
2
m

40


(20 + 20 + 10 + 10) m
s 

 
hs =  0.0083
+ 1.2 
= 111 m
3m

 2  9.81 m 


s2 

so that
kg  
m  π
m 


W s = γ Qhs = ρ gQhs = 1.23 3   9.81 2   (3 m)2  40
 111 m
s  
m 
s 4


= 3.79 ×105
N⋅m
= 379 kW
s
2(107)
8
107
4
6
8
0.00001
D
Problem 8.84
Air flows in a horizontal 100-ft-long, 24-in.× 24-in. duct at the rate of 5000 ft3/min. The air
then flows through an expansion into 200 ft of 36-in.× 36-in. duct. The expansion has a loss
coefficient of 0.80, based on the higher inlet velocity. Calculate the static pressure change
across the expansion and the static pressure loss across the entire duct system. The duct is
made of galvanized sheet metal. Use a constant air density of 0.0024 slug/ft3.
Solution 8.84
A2 = 36″ × 36″
A1 = 24″ × 24″
GALVANIZED SHEET METAL
V1
a
b
V2
c
100'
d
200'
Apply the mechanical energy equation from b to c to find (pb − pc).
pb
ρ
+
Vb2
p V2
+ g zb − ws = c + c + gzc + ghL .
2
ρ
2
Now ws = 0 and zb = zc. Then
pb − pc =
ρ
2
(Vc2 −Vb2 ) + ρ ghL
Using the continuity equation and
ghL = K L
V12 K LQ 2
=
2
2A12
gives
pb − pc =
ρQ 2  1
1 KL 
 2− 2+ 2
2  A2 A1 A1 
(
)(
)
2
2  min 
0.0024 slug / ft3 5000 ft3 / min 

 60 s   1 − 1 + 0.80 
pb − pc =


2
2
2
(2 ft)2 
 (3 ft) (2 ft)
 lb ⋅ s 2  ft 2 
×
 s1ug ⋅ ft 
 144 in.2 



or
pb – pc = 0.035 psi, a pressure drop form b to c.
Note that there is a static pressure drop since the friction losses give a pressure loss greater
than the static pressure rise due to the decrease in velocity. To find pa–pd, apply the
mechanical energy equation from a to d.
pa
ρ
+
2
Va2
p V
+ gza − ws = d + d + gzd + ghL .
ρ
2
2
Now ws = 0, za = zb. Then
V12
L1 V12
L2 V22 Q 2  K L
f1L1
f2 L2 
ghL = K L
+ f1
+ f2
=
+
 2 +
,
D1 2
D2 2
2
2  A1 D1A12 D2 A22 
and
Vd2 −Va2 Q 2  1
1 
=
 2 − 2 .
2
2  A2 A1 
The mechanical energy equation becomes
V 2 − Va2

pa − pd = ρ  d
+ ghL 
2


fL
f L 
ρQ 2  1
1 K
 2 − 2 + L2 + 1 12 + 2 22 
=
2  A2 A1 A1 D1A1 D 2 A2 


f2 L 2  1 
f1L1  
ρQ 2  1 
=
 2 1 +
 + 2  KL −1 +
 .
D2  A1 
D1  
2  A2 
The hydraulic diameters D1 and D2 are
D1 =
4A1 4(2 ft)2
=
= 2 ft and
4(2 ft)
p1
D2 =
4A2 4(3 ft)2
=
= 3 ft .
4(3 ft)
p2
Assuming standard air with v = 1.57 × 10−4 ft 2 /s, the Reynolds numbers are:
Re1 =
 min 
5000 ft3 / min ) (2 ft) 
(

V1D1 QD1
 60 s 
=
v
A1v
=
2
−4
2
(2 ft) (1.57 × 10 ft / s)
= 2.65 × 105
and
 min 
5000 ft3 / min ) ( 3ft ) 
(

V2 D 2 QD2
60 s 

Re2 =
=
=
= 1.77 ×105
2
−4 2
v
A2v
( 3ft ) (1.57 ×10 ft / s )
The absolute roughness for galvanized sheet metal (taken as galvanized iron) is found from
Table 8.1 to be ε = 0.0005 ft. Then
ε1 0.0005 ft
ε
0.0005 ft
=
= 0.00025 and 2 =
= 0.00017
D1
2 ft
D2
3 ft
The Moody chart then gives f1 = 0.0168 and f2 = 0.0172. Then
(
)(
2
)
2  min 
0.0024 slug / ft3 5000 ft3 / min 

 60s 
pa − pd =
2



1
0.0168(100 ft)    lb ⋅ s 2  ft 2 
 1  0.0172(200 ft) 
1+
0.80 − 1 +
,
×

+
  × 
2
2
2

(3 ft)
(2 ft)
 4 ft 2 
   slug ⋅ ft 
 144 in. 
 9 ft 2 


(
)
(
)
pa – pd = 0.00385 psi
or
 27.68 in.H 2 O 
pa − pd = ( 0.00385 psi ) 
,
psi


pa – pd = 0.107 in. H2O
Problem 8.85
The turbine shown in the figure below develops 400 kW. Determine the flowrate if (a) head
losses are negligible or (b) head loss due to friction in the pipe is considered. Assume
f = 0.02. Note: There may be more than one solution or there may be no solution to this
problem.
20 m
Diffuser
T
1m
120 m of 0.30-m-diameter
cast-iron pipe
Solution 8.85
(1)
•
20 m
Diffuser
•
(2)
1m
T
120 m of 0.30-m-diameter
cast-iron pipe
p1 V12
p V2
 V2
+
+ z1 = 2 + 2 + z2 + f
+ hT , where p1 = p2 = 0, z1 = 20 m, z2 = 0
D 2g
γ 2g
γ 2g
Thus, z1 =
V22
 V2
+f
+ hT
D 2g
2g
(1)
a) Neglect head losses ( f = 0):
N⋅m
400 × 103
52.0
P
V22
s
=
=
m
+ hT , where hT =
z1 =
2g
γQ 
V2
3 N π
2
 9.80 × 10
 (1 m) V2

m3  4
Thus,
20 m =
V22
52.0
m  V2

2  9.81 2 
s 

+
or
V23 − 392V2 + 1020 = 0
(2)
Determine the roots of this cubic equation. Let V23 − 392V2 + 1020 ≡ F
5000
4000
3000
2000
F
1000
0
–20
–10
–1000
0
10
20
–2000
–3000
V2
As indicated by the graph above, there at two real, positive roots to F = 0:
m
m
V2 = 2.65
V2 = 18.3
or
Thus,
s
s
Q = A2V2 =
Q = 2.08
π
4
(1 m)2V2 , or
m3
m3
or Q = 14.4
s
s
The negative root (V2 < 0) has no physical meaning.
b) Include head loss ( f = 0.02): From Eq. (1)
2
2
 1m 
V2 A2
 D2 
V=
= V2 
 = 11.1V2
 = V2 
A
 D 
 0.3 m 
or

 120 m 
V22
52.0
2
20 m = 1 + 0.02 
(11.1)
m
+


m
0.3
m
V




2

 2 9.81


s2 

Thus,
V23 − 0.398V2 + 1.034 = 0
Let G ≡ V23 − 0.398V2 + 1.034 ; determine V2 that gives G = 0.
1.8
1.6
1.4
1.2
G
1
0.8
0.6
0.4
0.2
0
–1
–0.5
0
0.5
1
V2
As indicated by the graph above, there is no positive real root.
Hence, the flow cannot occur with W s = 400 kW
Problem 8.86
Water flows from the nozzle attached to the spray tank shown in the figure below.
Determine the flowrate if the loss coefficient for the nozzle (based on upstream conditions)
is 0.75 and the friction factor for the rough hose is 0.11.
Nozzle diameter
= 7.5 mm
p = 150 kPa
D = 15 mm
𝓵 = 1.9 m
0.80 m
40°
Solution 8.86
Nozzle diameter
= 7.5 mm
p = 150 kPa
•
(2)
(1)
•
D = 15 mm
𝓵 = 1.9 m
0.80 m
40°
2
p1 V12
p V2
 
V
+
+ z1 = 2 + 2 + z2 +  f + K L 
,
γ 2g
γ 2g
 D
 2g
(1)
where p1 = 150 kPa, p2 = 0, z1 = 0.8 m, z2 = ℓ sin 40° = (1.9 m) sin 40° = 1.22 m, V1 = 0,
2
2
 D 
Q
 15 mm 
Q  A
V = , and V2 =
=
V = 
 V =
 V = 4V
A
A2  A2 
 7.5 mm 
 D2 
Thus, with f = 0.11 and KL = 0.75 Eq. (1) gives
N
V2
m 2 + 0.8 m = 1.22 m +  42 + 0.11 1.9 m  + 0.75 




N
 0.015 m 

 2  9.81 m 
9.80 ×103 3


m
s2 

or
m
V = 3.09
s
π
m
m3

Thus, Q = AV = (0.015 m)2  3.09  = 5.46 × 10 −4
4
s 
s

150 ×103
Problem 8.87
Water flows through the pipe shown in the figure below. Determine the net tension in the
bolts if minor losses are neglected and the wheels on which the pipe rests are frictionless.
3.0 m
Bolts
80 mm
Galvanized iron
10 m
20 m
Solution 8.87
(1)
•
3.0 m
Bolts
80 mm
Galvanized iron
(2)
•
•
(3)
10 m
FB = tension in bolts
p3A3
V3
20 m
(3)
V2
(2)
control volume
Application of the momentum equation to the control volume indicated gives
p3 A3 − FB = m (V2 − V3 ) = 0, since V2 = V3
Thus, FB = p3A3
(1)
Also,
p1 V12
p V2
 V2
, where p1 = p2 = 0, V1 = 0, z1 = 3 m, z2 = 0, and V2 = V
+
+ z1 = 2 + 2 + z2 + f
D 2g
γ 2g
γ 2g

 10 m + 20 m  
V2
 V 2

or 3 m = 1 + f 
Thus, z1 =  1 + f 

D  2g

 0.08 m   2  9.81 m 



s2 

m
or
58.9 = (1 + 375 f )V2, where V 
(2)
s
VD
(0.08 m)V
=
Also, Re =
or
Re = 7.14 × 104V,
(3)
2
v
−6 m
1.12 ×10
s
and from Table B.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7)
ε 0.15 mm
and Colebrook (Ref. 8)], =
= 0.00188
D
80 mm
Trial and error solution of Eqs. (2), (3), and (4)
for f, V, Re:
m
Assume f = 0.02; from (2) V = 2.63 ;
s
f
ε = 0.00188 (4)
D
Re
(4)
from (3) Re = 1.88 × 105; from (4) f = 0.0245 ≠ 0.02. Try again.
Assume f = 0.0245, or V = 2.40
Thus, V = 2.40
Obtain p3 from
or
p3 = f
m
, Re = 1.72 × 105 or f = 0.0245 Checks.
s
m
s
p3
γ
+

V32
p V2
V2
+ z3 = 2 + 2 + z2 + f 3−2
2g
γ 2g
D 2g
 3− 2 1
ρV 2
D 2
2
 20 m  1 
kg 
m
4 N
Hence, p3 = 0.0245 
  999 3  2.40  = 1.76 ×10
s 
m 
m2
 0.08 m  2 
Thus, from Eq. (1)
N  π 

FB =  1.76 × 10 4 2    (0.08 m)2 = 88.5 N
m  4 

Problem 8.88
When the pump shown in the figure below adds 0.2 hp to the flowing water, the pressures
indicated by the two gages are equal. Determine the flowrate.
Length of pipe between gages = 60 ft
Pipe diameter = 0.1 ft
Pipe friction factor = 0.03
Filter loss coefficient = 12
Filter
Pump
Filter
Pump
Solution 8.88
(2)
•
p1
γ
+
(1)
•
p
V12
V2
+ z1 + hp = 2 + 2 + z2 + hL
2g
γ
2g
As p1 = p2, z1 = z2 , V1 = V2
So, hp = hL
(1)
The pump adds 0.2 hp of power.
W = 0.2 hp ×
ft ⋅ lb
s = 110 ft ⋅ lb
1 hp
s
550
Convert to head by:
ft ⋅ lb
110

W
s = 1.76
hp =
=
lb
γ Q 62.4
Q
Q
3
ft
Sub into (1)
Q 
 
2
1.76  
V
 
 A
=  f +  KL 
=  f +  KL   
Q
 D
 2g  D
 2g
2
or
=
π
1.76 (2)( g )A2
Q =
where A = (0.1 ft)2 = 7.85 ×10 −3 ft 2
4
 

 f +  KL 
 D

3
1.76 (64.4)(7.85 ×10 −3 )2
60


+ 12 
 0.03
0.1


Q3 = 2.328 × 10−4
Q = 0.0615
ft3
s
Problem 8.89
The pump shown in the figure below adds 25 kW to the water and causes a flowrate of
m3
. Determine the flowrate expected if the pump is removed from the system. Assume
0.04
s
f = 0.016 for either case and neglect minor losses.
60-mm-diameter,
30-m-long pipe;
40-mm-diameter
f = 0.016
nozzle
Pump
Solution 8.89
(1)
•
60-mm-diameter,
30-m-long pipe;
40-mm-diameter
f = 0.016
nozzle
•
(2)
h
Pump
p1 V12
p V2
 V2
, where p1 = p2 = 0, z1 = h, z2 = 0, V1 = 0,
+
+ z1 + hp = 2 + 2 + z2 + f
D 2g
γ 2g
γ 2g
m3
m3
0.04
0.04
Q
m
Q
m
s
s
V2 =
=
= 31.8 , V = =
= 14.15
s
s
A2 π (0.04 m)2
A π (0.06 m)2
4
4
Thus,
2
2
m
m


14.15
 31.8



s 
s 
 30 m  
h + hP = 
+ 0.016 
= 133.2 m

m
m
0.06 m  


2  9.81 2 
2  9.81 2 
s 
s 


But,
N⋅m
25 ×103
P
s
hP =
=
= 63.8 m
3
γQ 

N
m

3

 9.80 ×10
  0.04
s 

m3  
Hence,
h = 133.2 m − 63.8 m = 69.5 m
Without the pump hp = 0 and z1 =
2
AV  D 
V2 =
=
 V or
A2  D2 
V22
 V2
where h = 69.5 m = z1 and
+f
D 2g
2g
2
 60 mm 
V2 = 
 V = 2.25V
 40 mm 
Thus,
 30 m  2
(2.25V )2 + 0.016 
V
0.06 m 

69.5 m =
m

2  9.81 2 
s 

so that
Q = AV =
π
V = 10.22
or
m
m

(0.06 m)2  10.22
 = 0.0289
4
s 
s

3
m
s
Problem 8.90
The vented storage tank shown in the figure below is used to refuel race cars at a race track.
A total of 40 ft of steel pipe (I.D. = 0.957 in.), two 90° regular elbows, and a globe valve
make up the system. Calculate the time needed to put 20 gal of fuel in a car tank. The
pressures, p2 and p1, are equal, the connections are threaded, and the fuel has the properties
of 68 °F normal octane ( ν = 8.31×10−6 ft2/s).
Vent
1
patm
Racing fuel
h = 12 ft
2
p2 = patm
Solution 8.90
Assume constant fluid density and apply the mechanical energy equation from point 1 to
point 2. No work is done.
p1 V12
p 2 V22
+
+ gz1 =
+
+ gz2 + ghL .
ρ
2
ρ
2
Now p1 = p2 and V1 ≈ 0. This gives
g ( z1 − z2 ) = ghL =
Since V2 =
4Q
π D2
2
V22 

V
+  K Lent + f + 2 K L 90° + K Lvalve  2 .
2 
D
 2
,
2


 8Q
g ( z1 − z2 ) =  1 + K Lent + f + 2 K L 90° + K Lvalve  2 4
D

π D
or
Q=
π D2
2 2
g ( z1 − z2 )



1 + K Lent + f + 2 K L 90° + K Lvalve 
D


(1)
Figure 8.22 gives KLent = 0.5, and Table 8.2 gives KL 90° = 1.5 and KLvalve = 10, the Reynolds
number is
VD 4Q
Re =
=
=
v
π Dv
4Q
(
 0.957 
ft  8.31×10 −6 ft 2 / s
π
 12

)
=
1.92 ×106 Q
( ft3 / s )
.
Table 8.1 gives
ε
D
=
0.00015 ft
= 0.0019.
 0.957 
ft 

 12

(2)
Using the known numerical values in Eq. (1) gives
Q=
 0.957 
ft 
 12

2 2
π
2
ft 

ft3
0.139
 32.2 2  (12 ft )
s 

s
=
, (3)
f ( 40 ft )
14.5 + 502 f
+ 2 (1.5 ) + 10
1 + 0.5 +
 0.957 
ft 

 12

where z1 – z2 is assumed constant for the refueling.
Equations (2) and (3) are now solved with the Moody chart for Q. Using a trial and error
procedure, guess f = 0.023 = wholly turbulent value. Then
Q=
0.139 ft3 / s
= 0.0272 ft3 / s,
14.5 + 502(0.023)
Re = 1.92 × 106 (0.0272) = 5.22 ×10 4 ,
and the Moody chart gives f = 0.0262. Then
Q=
0.139 ft3 / s
= 0.0264 ft3 / s,
14.5 + 502(0.0262)
Re = 1.92 ×106 ( 0.0264 ) = 5.07 ×104 ,
and the Moody chart gives f = 0.0262 so the iteration has converged. The time to put 20
gallons in a car is
t=
20 gal

ft3  
gal 
0.0264

  7.48 3 
s 
ft 

or
t = 101.3 s.
Problem 8.92
Calculate the water flowrate in the system shown in the figure below. The piping system
includes four gate valves, two half-open globe valves, fourteen 90° regular elbows, and 250
ft of 2-in. schedule 40 commercial steel pipe (with an actual inside diameter of 2.067 in.).
Assume threaded connections and a square-edged pipe entrance.
144 ft2
60 ft
220 ft
Piping system
20 ft
30 ft
72 ft2
Solution 8.92
Assume steady state, and constant density. Apply the mechanical energy equation from the
upper water surface 1 to the lower water surface 2,
p1 V12
p V2
+
+ gz1 − ws = 2 + 2 + gz2 + ghL
ρ
2
ρ
2
Now ws = 0 and p1 = p2 and assume V1 ≈ 0 and V2 ≈ 0. This gives ghL = g(z1 – z2)
2
 
V
where ghL =  f + 4 K L gate + 2 K L globe + 14 K L elb + K L ent + K L exit 
 D
 2
Substituting gives
1
2



2 g ( z1 − z2 )
V =


 f + 4 K Lgate + 2 K Lglobe + 14 K Lelb + K Lent + K Lexit 
 D

Table 8.2 and Fig. 8.22 give Kgate = 0.15, Kglobe = 2 ×10, Kelb = 1.5, Kent = 0.5 and K exit = 1.0
1

2


2


2 32.2 ft / s ( 220 ft )


V=
(250 ft)


 f  2.067  + 4(0.15) + 2(20) + 14(1.5) + 0.5 + 1.0 
 

ft 

  12

119 ft / s
=
1451 f + 63.1
(
)
Table 8.1 gives
ε
D
=
(0.00015 ft)
= 0.00087.
 2.067 
ft 

 12

Assuming 60°F water, the Reynolds number is
 2.067 
ft 
V
VD
12

 = (1.42 × 10 4 s/ ft)V .
=
Re =
5
2
−
v
1.21× 10 ft / s
The velocity V must be found by trial and error. Assuming wholly turbulent, the Moody
chart gives f = 0.019 so
V=
119 ft / s
= 12.5 ft / s,
1451(0.019) + 63.1
and
(
)
Re = 1.42 × 10 4 s /ft (12.5 ft/ s ) = 1.775 × 105.
The Moody chart gives f = 0.0207 so
V=
119 ft / s
= 12.3 ft / s
1451(0.0207) + 63.1
and
(
)
Re = 1.42 × 10 4 s /ft (12.3 ft/ s ) = 1.75 × 105
The Moody chart gives f = 0.0207 so convergence is obtained and
2
 π D2 
 π   2.067 
Q = VA = V 
ft 
= (12.3 ft / s)   
 4 
 4   12



or
ft 3
Q = 0.29
s
Problem 8.93
The pump shown in the figure below delivers a head of 250 ft to the water. Determine
the power that the pump adds to the water. The difference in elevation of the two ponds
is 200 ft.
KL
exit
= 1.0
Pump
KL
KL
valve
elbow
= 1.5
= 5.0
KL
ent
Pipe length = 500 ft
Pipe diameter = 0.75 ft
Pipe roughness = 0
= 0.8
Solution 8.93
KL
Pump
V
KL
KL
valve
exit
= 1.0
(2)
•
elbow
= 1.5
= 5.0
(1)
•
KL
ent
= 0.8
Pipe length = 500 ft
Pipe diameter = 0.75 ft
Pipe roughness = 0
V12
p
V2
− hL + hp = 2 + z2 + 2 , where p1 = p2 = 0, V1 = V2 = 0, z1 = 0, z2 = 200 ft,
γ
2g
γ
2g
hP = 250 ft
p1
+ z1 +
Thus,
 V2
V2
V2
V2
V2
= (0.8 + 4(1.5) + 5.0 + 1)
= 12.8
−f
−  K Li
+ hp = z2 so that with  K Li
D 2g i
2g
2g
2g
2g
i
  500 
 V2
−
f
−
12.8
  0.75 
 2(32.2) + 250 = 200

 

or
(1)
(667 f + 12.8)V2 = 3220
slugs 

1.94
V (0.75 ft)

3 
ρVD 
ft

=
Also, Re =
lb ⋅ s
μ
2.34 ×10 −5
ft 2
or
(2)
Re = 6.22 ×104V
And from the figure below:
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
4
6
104
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
107
ρ VD
Re = _____
μ
(1)
f
ε =0
D
Re
Trial and error solution. Assume
ft (2)
(1)
(3)
f = 0.02 ⎯⎯→V = 11.1 ⎯⎯→ Re = 6.9 ×105 ⎯⎯→ f = 0.012 ≠ 0.02
s
ft (2)
(1)
(3)
Assume f = 0.012 ⎯⎯→V = 12.4 ⎯⎯→ Re = 7.7 ×105 ⎯⎯→ f = 0.0121 ≈ 0.012
s
8
0.00001
ε
__
D
Thus, V = 12.4
ft
and
s
lb  π
ft 
ft ⋅ lb


W s = γ Qhp =  62.4 3  (0.75 ft)2  12.4  (250 ft) = 8.55 ×104
s
s
ft  4


ft ⋅ lb
hp
= 8.55 × 10 4
×1
= 155 hp
ft ⋅ lb
s
550
s
Alternatively, we could replace Eq. (3) (the Moody chart) by equation
 ε

 D
1
2.51 
= −2.0 log 
+
 (the Colebrook equation) and obtain V as follows.
3.7
f
Re
f




From Eq. (1),
1

2
3220
V =
 , which when combined with Eq. (2) gives
 (667 f + 12.8) 
(2)
4
1
2
3220
Re = 6.22 × 10 
 =
 (667 f + 12.8) 
3.53 × 106
1
(667 f + 12.8) 2
Also, the Colebrook equation with
(3)
ε
D
= 0 is
 2.51 
1
= −2.0 log 
 Re f 
f


By combining Eqs. (4) and (5), we obtain a single equation involving only f:
(4)
1

 2.51(667 f + 12.8) 2 
1
= −2.0 log 

6
f
 3.53 × 10 f



Using a compute root-finding program to solve Eq. (6) gives f = 0.0123, consistent with the
above trial and error method.
Problem 8.94
For the standpipe system shown in the figure below, calculate the flowrate for H = 4.0 ft, D
= 6.77 in., d = 0.125 in., and L = 48 in. The fluid is 70°F water. Assume steady flow and
neglect the energy loss in the entrance nozzle. The pipe is commercial steel.
D
(1)
H
(2)
L
d
Solution 8.94
Apply the mechanical energy equation from point 1 to point 2. Assume constant density.
p1 V12
p V2
+
+ gz1 = 2 + 2 + g z2 + ghL .
ρ
2
ρ
2
Now p1 = p2, V1 ≈ 0, and z1 – z2 = H. The energy equation is
ghL = gH −
V22
.
2
Neglecting the energy loss in the entrance nozzle,
ghL = f
L V22
,
d 2
f
V2
L V22
= gH − 2 ,
2
d 2
or
V2 =
2 gH
.
L
1+ f
d
This is a velocity flowrate problem.
We will solve for V2 using the above equation and the Moody chart. For commercial steel
pipe, the Table 8.1 gives
ε
D
=
0.00015 ft
= 0.0144.
 0.125 
ft 

 12

We first choose the Moody friction factor for wholly turbulent flow, f = 0.044. Then
V2 =
(
)
2 32.2 ft / s 2 (4 ft)
 48 in. 
1 + 0.044 

 0.124 in. 
= 3.79 ft / s.
Using Table B.1, v = 1.052 × 10−5 ft2/s and
Vd
Re = 2 =
v
 0.125 
(3.79 ft / s) 
ft 
 12
 3750.
−5 2
1.052 × 10 ft / s
(
)
The Moody chart gives f = 0.053 so
V2 =
(
)
2 32.2 ft / s2 (4 ft)
 48 in. 
1 + 0.053 

 0.125 in. 
= 3.47 ft / s.
Then
 0.125 
(3.47 ft / s) 
ft 
V2 d
12

 3436.
Re =
=
−
5
2
v
1.052 × 10 ft / s
(
)
And the Moody chart gives f = 0.054
Another iteration indicates convergence has been obtained,
 0.125 
(3.47ft / s)π 
ft 
2
πd
12


Q = V2
=
4
4
Q = 2.96 ×10 −4
ft3
.
s
2
Problem 8.95
Water flows through two sections of the vertical pipe shown in the figure below. The
bellows connection cannot support any force in the vertical direction. The 0.4-ft-diameter
lb
pipe weighs 0.2 , and the friction factor is assumed to be 0.02. At what velocity will the
ft
force, F, required to hold the pipe be zero?
Free jet
F
f = 0.020
Pipe weighs
0.20 lb/ft
D = 0.40 ft
Bellows
V
Solution 8.95
V2
(2)
WH2O
ℓ
Wpipe
(1)
V1
p1A1
From the momentum equation applied to the control volume indicated
p1A1 −WH2O −Wpipe = m (V2 −V1 ) = 0 since V1 = V2
Thus, p1 =
WH 2O +W pipe
A1
 W pipe 

  
γ A1 +  
=
A1
or
lb 

 0.20  
1b
ft 
= γ  + 1.59, where p1 ~ 2 , ℓ ~ ft
p1 = γ  + 
π
ft
(0.4 ft)2
4
Also,
p1 V12
p V2
 V2
, where p2 = 0, V1 = V2 = V, z1 = 0, and z2 = ℓ
+
+ z1 = 2 + 2 + z2 + f
D 2g
γ 2g
γ 2g
Thus,
p1 = γ z2 + f
 1
ρV 2 ,
D2
or when combined with the above force balance result
p1 = γ  + f
That is,
 1
ρV 2 = γ  + 1.59
D2
2D(1.59)
2(0.4)(1.59)
ft
f ρV 2
=
= 5.73
= 1.59 or V =
ρf
(1.94)(0.02)
s
2D
Note: This answer is independent of the pipe length, ℓ.
Problem 8.96
Water is circulated from a large tank, through a filter, and back to the tank as shown in the
figure below. The power added to the water by the pump is 200 ft ∙ lb/s. Determine the
flowrate through the filter.
KL exit = 1.0
KL elbow = 1.5
KL valve = 6.0
KL ent = 0.8
KL filter = 12.0
Filter
200 ft. of 0.1-ft-diameter
pipe with ε/D = 0.01
Pump
Solution 8.96
p1
γ
+ z1 +
V 2
V12
p
V2  
+ hP = 2 + z2 + 2 +  f +  K Ll 
 2g
2g
γ
2 g  D i

(1)
where p1 = p2, V1 = V2 = 0, and z1 = z2
Also, W = γ Qh or
P
P
200
hP =
62.4
ft ⋅ lb
s
lb  π

(0.1 ft)2 V
34

ft 
=
408
V
Thus, Eq. (1) becomes

408  200 ft
V2
=
f + (0.8 + 5(1.5) + 12 + 6 + 1) 
V
 0.1 ft
 2  32.2 ft 


s2 

or
V3 =
13.13
( f + 0.01365)
(2)
Also,
Re =
ρVD
=
μ
1.94
slugs  ft 
V  (0.1 ft)
ft3  s 
or
lb ⋅ s
2.34 × 10 −5 2
ft
Re = 8290V
(3)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
6
2(104)
8
103
104
4
6
2(105)
8
4
105
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Trial and error solution:
ft
Assume f = 0.04. From Eq. (2), V = 6.26 ; from Eq. (3), Re = 5.20 × 104. Thus, from the
s
figure above, f = 0.039 ≠ 0.04
Assume f = 0.039, or V = 6.29
Thus, Q = AV =
ft
and Re = 5.21 × 104 and f = 0.039 (Checks)
s
π
ft 
ft3

(0.1 ft)2  6.29  = 0.0494
4
s
s

Alternatively, the Colebrook equation could be used rather than the Moody chart. Thus,
 ε

 D
1
2.51 
= −2.0 log 
+
, where from Eq. (2)
f
 3.7 Re f 


(4)
 13.13 
f =  3  − 0.01365
 V 
(5)
ε
__
D
Thus, by combining Eqs. (3), (4), and (5), we obtain the following equation for V:




0.01
+ 2.51


1
3.7


2.0
log
=
−
1
1


 13.13 
2
  13.13 
2 

(8290V )   3  − 0.01365 
 3  − 0.01365 

 V 

 V 
 
(6)
Using a computer root-finding program gives the solution to Eq. (6) as V = 6.29
same as obtained by the above trial and error method.
ft
, the
s
Problem 8.97
A thief siphoned 15 gal of gasoline from a gas tank in the middle of the night. The gas
tank is 12 in. wide, 24 in. long, and 18 in. high and was full when the thief started. The
siphoning plastic tube has an inside diameter of 0.5 in. and a length of 4.0 ft. Assume
that at any instant of time, the steady-state mechanical energy equation is adequate to
predict the gasoline flowrate through the tube. As 15 gal is 3465 in.3, the gasoline level in
the tank will drop 12.0 in. You may use the gasoline level after it has dropped 6.0 in. to
estimate the average gasoline flow rate. Use this flowrate to estimate the time needed to
siphon the 15 gal of gasoline. Compare your answer with the answer of 190 s found in
problem 3.107, using Bernoulli’s equation. The siphon discharges at the level of the bottom
of the gasoline tank. You may find it useful to use the Blasius equation for smooth pipes
found in problem 8.45.
Solution 8.97
Assume constant fluid density and apply the mechanical energy equation from the gasoline
surface (1) to the siphon tube discharge (2).
1
h
2
V
p1 V12
p V2
+
+ gz1 − ws = 2 + 2 + gz2 + ghL .
ρ
ρ
2
2
Now p1 = p2 = patm, V1 = 0, ws = 0, and z1 – z2 = h .This gives
gh =
V22
+ ghL
2
(1)
where
  V22

.
ghL =  K L ent + f 
D 2

(2)
Figure 8.22 gives KL ent = 0.8. Assuming a smooth pipe (see Table 8.1), the Blasius equation
for smooth pipes and 4000 < Re < 105 gives
1
1
1
−
4


4
Q
π
Dv
4


= 0.3164
f = 0.3164 Re 4 = 0.3164
)
−


 π Dv 


 4Q 
(3)
Equations (1), (2), and (3) give
1

2
4  8Q



π
Dv

gh = 1 + K Lent + 0.3164 
.

D  4Q   π 2 D 4


Substituting the known numerical values, and using Table 1.5 gives
1

2  4 
  1 

ft
 π  ft   4.9 × 10 −6
 

s   
0.3164 ( 4.0 ft )   24  
ft 
8Q 2


=
+
+
×
32.2
1.0
ft
1
0.9
,
) 

(

 
4
4Q
 1 
s2 

1

  π 2  ft 

 ft 


24

 



 24 




1 

3

ft
0.6078(ft / s) 4  2.69 × 105 Q 2
32.2 2 = 1.9 +
,

1
s
ft 4


Q4


2
or
2
(
Q + 0.3199 ft
3
1
/ s 4 Q1.75 − 6.30 ×10 −5 ft 6 / s2 = 0.
)
A trial and error approach produced a single, positive root of Q = 5.374 × 10−3 ft3/s. A
check on the Reynolds number gives
4Q
Re =
=
π Dv
(
4 5.374 × 10 −3 ft 3 / s
)
 1 
ft  (4.9 × 10 −6 ft 2 / s)
 24 
π
= 3.35 × 10 4
or turbulent as assumed and within the 4000 to 105 range for Eq. (3). The time to drain 15
gallons of gasoline is
 ft3  
s

t = 15 gal 


 7.48ga1   5.374 ×10−3 ft3 


or
t = 373s = 6 min,13s compared to t = 190s = 3min,10s using Bernoulli’s equation.
Problem 8.98
Estimate the time required for the water depth in the reservoir shown in the figure below to
drop from a height of 25 m to 5 m. The connections are threaded.
500 m2
T = 10°C
25 m
5m
30 m
1000 m of 7-cm-inside-diameter,
galvanized iron pipe, five regular
90° elbows, one fully open globe valve
Solution 8.98
500 m2
1
•
T = 10°C
25 m
5m
30 m
1000 m of 7-cm-inside-diameter,
galvanized iron pipe, five standard
90° elbows, one fully open globe valve
•
2
Assume quasi-steady state at each instant of time and constant fluid density and apply the
mechanical energy equation from the water surface 1 to the pipe outlet 2,
p1 V12
p V2
+
+ gz1 − ws = 2 + 2 + gz2 + ghL .
ρ
ρ
2
2
V2 2
Now p1 = p2 = patm and ws = 0. Assume V1 ≈ 0. This gives g ( z1 − z2 ) =
+ ghL
2
where
2
 
V
ghL =  f + 5K L elb + K Lglobe + K Lreent  2
 D
 2
so
2


 8Q
g ( z1 − z2 ) =  1 + f + 5 K Lelb + K Lglobe + K Lreent  2 4
D

π D
or
Q = π D2
g ( z1 − z2 )



8 1 + f + 5K L elb + K L globe + K L reent 
D


Denote (z1 − z2) by (h + 30m). Table 8.2 and Fig. 8.22 give
K L elb = 1.5, K L globe = 10, and K L reent = 0.8.
Then
Q = π (0.07 m)
2
(9.81 m/s2 ) ( h + 30 m)


 1000 m 
8 1 + f 
 + 5(1.5) + 10 + 0.8
 0.07 m 


m 2.5
h + 30 m
= 0.0170
s
193 + 14300 f
Using Table B.2, the Reynolds number is
Re =
V2 D 4Q
4Q
=
=
= 1.39 × 107 Q
6
2
−
v
π Dv π (0.07 m)(1.31×10 m / s)
where Q is in m3/s. Table 8.1 gives


m
(0.15 mm) 

1000 mm 
ε

=
= 0.00214.
D
(0.07 m)
Assuming wholly turbulent flow, the Moody chart gives f = 0.0236.
Then


m 2.5 
h + 30 m
m 2.5 
Q =  0.0170
=  0.000900

 h + 30 m ,

s  19.3 + 14300(0.0236) 
s 

Evaluating Q for h = 25 m and h = 5 m gives
Q25 = 0.00667 m3/s and Q5 = 0.00532 m3/s.
The corresponding Reynolds numbers are
R25 = (1.39 × 107 )(0.00667) = 9.27 × 10 4
and
R5 = (1.39 × 10 7 )(0.00532) = 7.40 × 10 4.
Therefore, the flow is turbulent. The Moody chart gives f25 = 0.0256 and f5 = 0.026.
Repeating the calculations for the average f25 = f5 = 0.0258 gives

m 2.5
h + 30 m
m 2.5 
Q = 0.0170
=  0.000863
 h + 30 m
s
19.3 + 14300(0.0258) 
s 

m 2.5 
m3
Q25 =  0.000863
55 m = 0.00640
,



s
s



m 2.5 
m3
Q5 =  0.000863
35
m
0.00511
,
=


s
s


R25 = (1.39 ×107 )(0.00640) = 8.90 × 10 4 ,
R5 = (1.39 × 107 )(0.00511) = 7.10 × 10 4 ,
f25 ≈ 0.0256 and f5 ≈ 0.026
With no perceptible change, use f ≈ 0.0258

m 2.5 
h + 30 m
Q =  0.0170


s  19.3 + 14300(0.0258)


m2.5 
Q =  0.000863
 h + 30 m.

s


(1)
Applying the conservation of mass to the tank gives
dh
Q = −A
(2)
dt
where A = 500 m2. Eliminating Q from Eqs. (1) and (2) gives

m 2.5 
dh
 0.000863
 h + 30 m = −A ,
s 
dt

t
5m
dh
−A
,
0 dt = 

2.5  25 m h + 30 m
m
 0.000863

s 

−A(2)
t=
[ h + 30 m ]525mm
2.5

m 
 0.000863

s 

=
−1000 m 2

m 2.5 
 0.000863

s 

( 35 m − 55 m )
 hr 
= 1.74 × 106 s 

 3600 s 
t = 483 hrs.
We were fortunate in this problem in that approximately the same friction factor was
appropriate for all water depths. The solution would have been much more difficult if this
were not the case.
Problem 8.99
Sheldon and Leonard come home from a long day of studying to discover that their
basement is flooded with water. They have a submersible pump for such an emergency and
connect the pump discharge to a 12-ft-long garden hose, as shown in the figure below. The
garden hose may be considered a smooth plastic pipe with a 1.0-in. inside diameter. If the
initial water depth is 6.0 in. and the pump has an input power of 0.25 hp and an overall
efficiency of 0.60, how long will it take to pump out the basement? The pump inlet piping
length is negligible, and the garden hose outlet is 7.0 ft above the basement floor.
Q
h0 = 6 in.
24 ft
h = 7 ft
36 ft
Solution 8.99
Assume constant water density and apply the mechanical energy equation to a streamline
connecting the water surface (1) and the hose outlet (2).
p1 V12
p V2
+
+ gz1 + w p = 2 + 2 + gz2 + ghL
ρ
2
ρ
2
Now p1 = p2 = patm and assume V1 ≈ 0. This gives
V22
+ ghL
g ( z1 − z2 ) + w p =
2
Now ghL accounts for the mechanical energy loss in the inlet and discharge pipes. Assume a
re-entrant inlet with a 1.0 in. inlet pipe to get
 V 2

ghL =  K L reent + f  2 .
D 2

Figure 8.22 gives KL reent = 0.8. The energy equation is
  V22

g ( z1 − z2 ) + w p =  1 + K L reent + f 
,
D 2

W p 
  8Q 2
g ( z1 − z2 ) +
= 1 + K L reent + f  2 4 ,
m
D π D

W p 
  8Q 2
g ( z1 − z2 ) +
= 1 + K L reent + f  2 4 ,
ρQ 
D π D
or
 3
8ρ 

+
K
+
f
1
L
reent

 Q − ρ g ( z1 − z2 )Q −W p = 0.
2 4
D
π D 
We have two ways to solve for Q. We could solve for Q(t) for each water depth in the
basement or we could find Q for the overage water depth. We will choose the second where
z2 − z1 = 6.75 ft. Assuming 60°F water, the numerical values give

8ρ 
1 + K L reent + f  Q3
2 4
D
π D 
lbm  


8  62.4 3  

 3
lb ⋅ s 2
ft  1 + 0.8 + f (12 ft)  
= 
Q

4
 32.2 ft ⋅ lbm 
1




1



 ft   
π 2  ft  
12




 12 
= 32570(1.8 + 144 f )
lb ⋅ s 2
8
Q3 ,
ft
1b 
lb

ρ g ( z1 − z2 )Q =  62.4 3  ( −6.75 ft)Q = 421.2 2 Q
ft 
ft

and noting that W p represents the (useful) mechanical power transferred to the fluid,
 550 ft ⋅ lb 
ft ⋅ lb
W p = 0.60 ( 0.25 hp ) 
 = 82.5
s
 s ⋅ hp 
to give
 lb ⋅ s2  3 
lb 
ft ⋅ lb
32570(1.8 + 144 f )  8  Q −  421.2 2  Q − 82.5
=0
 ft 
s
ft 



or
Q=3
0.0129  ft6 
0.00253 ft9
.

Q +
(1.8 + 144 f )  s2 
(1.8 + 144 f ) s3
(1)
We now solve for Q by trial and error. However, we first develop an expression for the
Reynolds number using Table B.1.
VD 4Q
Re =
=
v
π Dv
4Q
=
 1 
π  ft  (1.21×10 −5 ft 2 / s)
 12 
s 

= 1.26 × 106 3  Q.
ft 

ft 3
Guess an initial Q = 0.1
s
5
Re = 1.26 × 10 .
The Moody chart gives f = 0.0172 so Eq. (1) gives Q = 0.0959 ft3/s.
Next guess Q = 0.0959 ft3/s so R = 1.21 × 105.
The Moody chart gives f = 0.0173, so with one more iteration convergence is obtained and
Q = 0.0957 ft3/s.
The time to pump out the basement is
1 
ft (36 ft)(24 ft)
Volume  2 
=
t=
,
Q
(0.0957 ft3 / s)
 min 
t = 4514 s 
,
 60 s 
or
t = 75.2 min.
Problem 8.100
A company markets ethylene glycol antifreeze in half-gallon bottles. A machine fills and
caps the bottles at a rate of 60 per minute. The 68 °F ethylene glycol (ρ = 69.3 lbm /ft3,
v = 1.93 × 10−4 ft2/s) is pumped to the machine from a tank 36 ft away. The pressure at the
discharge of the pump is 35 psig, and the pressure at the inlet to the filling machine must be
at least 15 psig. Determine the diameter necessary for drawn copper pipe. The pipe has no
elevation change.
Solution 8.100
Assume constant fluid density and apply the mechanical energy equation to the pipeline
from the inlet (i) to the outlet (0) with no work.
Vi 2
po Vo2
+
+ gzi =
+
+ gzo + ghL .
ρ
ρ
2
2
pi
The continuity equation gives Vi = Vo. The problem statement gives zi = zo so
ghL =
V=
(35 − 15)
pi − po
Q
=
A
ghL = f
=
ρ
lb 
in.2 
144


in.2 
ft 2 

lbm  
lb ⋅ s

 69.3 3   32.2 ft ⋅ lbm 
ft  


2
gal  1 ft3 
0.5


s  7.48 gal 
π 2
D
4
=
0.0851
= 1338 ft 2 / s2 .
ft3
s
D2
 V2
D 2
2
ft 2
1 36 ft  0.0851  ft 2
1338 2 = f


2
D  D 2  s2
s
1
 f 5
D=

 10264 
Re =
4Q
=
π Dv
or
1
5
D = 0.158 f
gal   1 ft 3 

4  0.5


s   7.48 gal 


π D 1.93 × 10 −4
ft 2 

s 
=
441
D

Guess f = 0.02, so D = 0.072 and Re = 6100.
The Moody chart gives f = 0.0353.
Using this new f, D = 0.081 and Re = 5448.
The Moody chart gives f = 0.0365.
Again D = 0.081, so the solution has converged.
D = 0.081 ft = 0.97 in.
Problem 8.101
A certain process requires 2.3 cfs of water to be delivered at a pressure of 30 psi. This water
comes from a large-diameter supply main in which the pressure remains at 60 psi. If the
galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded
90° elbows, determine the pipe diameter. Elevation differences are negligible.
Solution 8.101
•
(1)
Q
(2)
D
2
p1 V12
p V2
 
V
+
+ z1 = 2 + 2 + z2 +  f +  K L 
, where p2 = 30 psi, p1 = 60 psi, z1 = z2, V1 =
γ 2g
γ 2g
 D
 2g
ft3
Q
s = 2.93 ft , with D ~ ft
0, V2 = V = =
A π D2
D2 s
4
2.3
Thus,
 
1
p1 − p2 =  f +  K L  ρV 2
 D
2
or
lb 
in.2 
s1ugs 
 200 ft 
 2.93 ft   1  
(60 − 30) 2 144 2  = (1 + f 
 + 6(1.5) + 0.5)  2   1.94 3 


in. 
ft 
ft 
 D 
 D s   2 
2
where we have used
 KL = 6 KL
elbow
+  KLentrance = 6(1.5) + 0.5
Thus,
 19.0 f  1
49.4 = 1 +

D  D4

(1)
Also,
 2.93 
D
VD  D 2 
Re =
=
=
v
v
ft 2
s
ft 2
1.21× 10 −5
D
s
2.93
or Re = 2.42 × 105
1
D
(2)
and from Table B.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7)
and Colebrook (Ref. 8)]
ε
D
=
0.0005 ft
D
(3)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
4
6
2(106)
8
105
4
106
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Finally, from the figure above:
f
(4)
D
ε
, and Re.
D
Normally, it is easiest to guess a value of f, calculate D, etc. In this case (because of minor
losses), Eq. (1) is not easy to use in this fashion. Thus, assume D, calculate f [Eq. (1)], Re
Trial and error solution of Eqs. (1), (2), (3), and (4) for f, D,
[Eq. (2)], and
ε
D
[Eq. (3)]. Look up f in the figure above [Eq. (4)] and compare with that
from Eq. (1).
Assume D = 0.4 ft. Thus, f = 0.00557, Re = 6.05 × 105,
above f = 0.021 ≠ 0.00557
ε
D
ε
= 0.00125 or from the figure
= 0.001 or f = 0.0203 ≠ 0.0551
D
ε
Assume D = 0.45 ft. Thus, f = 0.0243, Re = 5.38 × 105,
= 0.00111 or f = 0.0205 ≠ 0.0243
D
ε
Assume D = 0.44 ft. Thus, f = 0.0197, Re = 5.50 × 105, = 0.00114 or f = 0.0205 ≠ 0.0197
D
After enough trial obtain D = 0.442 ft
Assume D = 0.5 ft. Thus, f = 0.0551, Re = 4.84 × 105,
Note: If the figure above [Eq. (4)] is replaced by the Colebrook equation, this problem can
be solved as follows.
Thus, from Eq. (1),
 ε

 D
1
2.51 
(49.4D5 − D )
= −2.0 log 
+
f =
so that with the Colebrook equation
, when
3.7
19
f
Re
f




combined with Eqs. (2) and (3), gives
1



2
 0.0005

19
2.51D 19
=
−
+
2.0
log



5
1
 (49.4D − D ) 
 3.7D 2.42 × 105 (49.4D5 − D ) 2 


(5)
Using a computer root-finding routine gives the solution to Eq. (5) as D = 0.442 ft which is
the same as that obtained by the trial and error method above.
Problem 8.102
Water is pumped between two large open reservoirs through 1. 5 km of smooth pipe. The
water surfaces in the two reservoirs are at the same elevation. When the pump adds 20 kW
m3
to the water, the flowrate is 1
. If minor losses are negligible, determine the pipe
s
diameter.
Solution 8.102
(1)
•
(2)
•
V
PUMP
V2
p
V2
+ z1 + 1 + hs − hL = 2 + z2 + 2 , where p1 = p2 = 0, V1 = V2 = 0, z1 = z 2
γ
2g
γ
2g
Thus,
N⋅m
20 × 103

W
s
(1) hs = hL where hs = s =
= 2.04 m
3
γQ 
3 N  m 

 9.80 × 10
 1

m3   s 
p1
and
hL = f
2
V
with
D 2g
m3
Q
s = 1.273 m with D  m
V= =
A π D2
D2 s
4
1
Hence,
2
2
 1.23  m
f
1.5 × 103 m  D 2  s 2
(2) hL = f
= 123.9 5 m
m
D

D
2  9.81 2 
s 

From Eqs. (1) and (2), 2.04 = 123.9
1
(3) D = 2.27 f 5
f
D
5
or f = 0.0165D5
Also,
Re =
(4) Re =
ρVD
=
μ
kg  1.273 

 m Dm
m3  D 2 
or
−3 N ⋅ s
1.l2 × 10
m2
999
1.14 × 106
D
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
6
2(104)
8
103
4
104
Finally, with
ε
D
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
= 0, the Moody chart (the figure above) is the final equation.
(5)
f
ε =0
D
Re
Trial and error solution of Eqs. (3), (4), and (5) for f, Re,
and D:
1
Assume f = 0.02, so Eq. (3) gives D = 2.27 ( 0.02 ) 5 = 1.04 m and Eq. (4) gives
1.14. ×106
Re =
= 1.10 ×106. Thus, from Eq. (5), f = 0.0115, which is not equal to the
1.04
assumed f = 0.02. Try again with f = 0.0115, which gives D = 0.931 m, Re = 1.22 ×
104, and f = 0.0113 ≠ 0.0115. One final try with f = 0.0113 gives D = 0.927 m, Re =
1.23 × 104, and f = 0.0113 as assumed. Thus, D = 0.927 m.
 ε

 D
1
2.51 
= −2.0 log 
+
An alternate method is to use the Colebrook formula

f
 3.7 Re f 


rather than the Moody chart [Eq. (5)]. Thus, with
ε
D
= 0, the Colebrook formula is
 2.51 
1
= −2.0 log 
which, when combined with Eqs. (3) and (4), gives
 Re f 
f


(6)




2.51D
= −2.0 log 
1
1
1.14 ×106 (0.0165D5 ) 2 
(0.0165D5 ) 2

 jkljkj
1
Using a computer root-finding program to solve Eq. (6) gives D = 0.926, which is
consistent with the trial and error solution given above.
Problem 8.103
Determine the diameter of a steel pipe that is to carry 2000
gal
of gasoline with a pressure
min
drop of 5 psi per 100 ft of horizontal pipe.
Solution 8.103
p1 V12
p V2
 V2
+
+ z1 = 2 + 2 + z2 + f
, where z1 = z2, and V1 = V2
D 2g
γ 2g
γ 2g
Thus,
 1
p1 − p2 = f
ρV 2
D2
ρ = 1.32
V=
Q
=
A
s1ugs
ft
3
lb 
in.2 
p1 − p2 = 5 2 144 2  ,  = 100 ft,
in. 
ft 
with
, μ = 6.5 × 10 −6
lb ⋅ s
ft 2
, and
gal   1min  
in.2   1 ft3 

2000
231





min   60 s  
gal   1728 in.3 

π
4
(1)
D2
, or V =
5.67 ft
with D ~ ft
D2 s
Hence, Eq. (1) gives:
5(144)
 100 ft  1 
slug   5.67 ft 
= f
  1.32 3   2

2
ft
ft   D s 
 D ft  2 
lb
2
or
1
5
D = 1.24 f
(2)
slugs   5.67 ft 

1.32 3   2
 D ft

1.15 × 106
ρVD 
s
ft
D


Also, Re =
, or Re =
=
lb ⋅ s
D
μ
6.5 × 10 −6 2
ft
(3)
and
ε
0.00015
, where D ~ ft
(4)
D
D
Finally, the fourth equation is the Moody chart (or the Colebrook equation)
=
(5)
f
Re
Note: Four equations [Eqs. (2), (3), (4), and (5)] and four unknowns ( f,
ε
D
, D, and Re)
Trial and error solution:


(5) f = 0.0148 ≠ 0.02
Guess f = 0.02 ⎯⎯→ D = 0.567 ft
ε
(4) 
= 0.000265

D

(3)  Re = 2.03 ×106
(2)
Thus, the guessed value is not correct.


(5)  f = 0.0150 ≈ 0.0148
Guess f = 0.0148 ⎯⎯→ D = 0.534 ft
ε
= 0.000281
(4) 


D
(2)
(3)  Re = 2.15 ×106
1
Thus, D = 1.24(0.0150)5 = 0.535 ft
Using the Colebrook equation, rather than the Moody chart, Eq. (5), we have
 ε

 D
1
2.51 
= −2.0 log 
+
 which, using Eqs (2), (3), and (4) is,
f
 3.7 Re f 






 0.0015

1
2.51D
= −2.0 log 
+

5
5
3.7D

D
D

2
2 
6
×
1.15
10




 
 1.24 
 1.24  

(6)
Using a computer root-finding program to solve Eq. (6) gives D = 0.563 ft which is
consistent with the above trial and error solution.
Problem 8.104
Water is to be moved from a large, closed tank in which the air pressure is 20 psi into a
ft 3
large, open tank through 2000 ft of smooth pipe at the rate of 3
. The fluid level in the
s
open tank is 150 ft below that in the closed tank. Determine the required diameter of
the pipe. Neglect minor losses.
Solution 8.104
(1)
•
150 ft
D
V
(2)
•
ℓ = 2000 ft
V2 p
V2
 V2
+ z1 + 1 = 2 + z2 + 2 + f
,
D 2g
γ
2g γ
2g
p1
(1)
where V1 = V2 = 0, z1 = z2 = 150 ft, and p1 = 20 ft, p2 = 0
Also,
ft3
3
Q
3.82
ft
V = = s = 2 , where V ~ , D ~ ft
A π D2 D
s
4
Thus, Eq. (1) becomes
2
lb  
in.2 

 3.82 
20
144




ft 2 
 in.2  
2000 ft  D 2 
+ 150 ft = f
lb
ft 
D

62.4 3
2  32.2 2 
ft
s 

or
1
5
D = 1.18 f
(2)
Also,
Re =
ρVD
=
μ
 3.82 
D
5
 D 2  = 1.94(3.82) , or Re = 3.17 ×10
μ
D
2.34 ×10 −5 D
ρ
(3)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
6
2(104)
8
103
104
4
6
2(105)
8
105
4
6
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Trial and error solution:
Assume f = 0.02 so from Eq. (2), D = 0.540 ft and from Eq. (3), Re = 5.87 × 105. Thus, from
ε
= 0) f = 0.013 ≠ 0.02
D
Assume f = 0.013, which gives D = 0.495 ft, Re = 6.40 × 105, and f = 0.012
the figure above (with
Assume f = 0.0125, so D = 0.491 ft, Re = 6.46 × 105, f = 0.0125 (Checks)
Thus, D = 0.491 ft
 ε


1
2.51 
= −2.0 log  D +
Alternately, the Colebrook equation,
, rather than the Moody
3.7
f
Re
f




chart, the figure above, could be used as follows:
With
ε
D
= 0, the Colebrook equation is
 2.51 
1
= −2.0 log 
where
 Re f 
f


(4)
5
 D 
From Eq. (2), f = 

 1.18 
(5)
Thus, combining Eqs. (3), (4), and (5) gives






1
2.51


2.0
log
=
−
5
5


5
3.17 × 10  D  2 
 D 2


 



 
D
 1.18 
  1.18  

(6)
Using a computer root-finding technique gives the solution to Eq. (6) as D = 0.492 ft, which
is consistent with the above trial and error solution.
Problem 8.105
A commercial steel flow channel in a heat exchanger has an equilateral triangle cross
section with each side measuring 5.0 in. and a length measuring 96.0 in. Water at 60 °F
flowing through the channel has a pressure loss of 0.10 psi. Find the water flowrate.
b
Solution 8.105
The pressure loss is
h
fL V 2
ΔpL = ρ ghL = ρ
.
Dh 2
b
bh b 2 b2
3 2
A=
=
b −
=
b
2 2
4
4
Dh =
V=
4A
=
P
3 2
b
3b
b
4
=
=
3b
3
3
4
Q
4Q
=
A
3b2
Substituting,
ΔpL =
ρ fL  3   4Q 
2


 =
2  b   3b2 
8 ρ fLQ 2
3b5
or
Q=
3b5
ΔpL.
8ρ fL
(1)
Assuming the flow is turbulent, calculate the relative roughness. Using Table 8.1,
ε
Dh
=
3ε
=
b
3(0.00015 ft)
= 0.000624.
 ft 
(5.0 in.) 

 12 in. 
Equation (1) gives
Q2 =
5
lb  
in.2 
 5.0  
3
ft   0.10 2  144 2 
 12  
in.  
ft 
slugs   96   lb ⋅ s2 

8 1.94 3  f  ft  

ft   12   slug ⋅ ft 

or
Q=
0.0502 ft3 / s
.
f
=
0.00252 ft 6
f
s2
The Reynolds number is
4Q
4Q(ft3 / s)
Re =
=
= 2.64 × 105 Q.
2
3bv
ft 
 5 
3  ft  1.21×10 −5

s 
 12  
We now guess Q = 0.375 ft3/s. Then
Re = 2.64 ×105 ( 0.375 ) = 9.90 ×104.
The Moody chart gives f = 0.0208 so
Q=
0.0502 ft3 / s
= 0.349 ft3 / s.
0.0208
Our second guess is Q = 0.375 ft3/s. Then,
Re = 2.64 ×105 (0.345) = 9.11×104.
The Moody chart gives f = 0.0211 so
Q=
0.0502 ft3 / s
= 0.346 ft3 / s.
0.0211
This is close enough to our previous guess so
Q = 0.346
ft3
.
s
Problem 8.106
Rainwater flows through the galvanized iron downspout shown in the figure below at a rate
m3
of 0.006
. Determine the size of the downspout cross section if it is a rectangle with an
s
aspect ratio of 1.7 to 1 and it is completely filled with water. Neglect the velocity of the
water in the gutter at the free surface and the head loss associated with the elbow.
70 mm
g
4m
3m
Solution 8.106
p1 V12
p V2
 V2
+
+ z1 = 2 + 2 + z2 + f
, where p1 = p2 = 0, V1 = 0, V2 = V1, z1 = 4.07 m ,
γ 2g
γ 2g
Dh 2 g
and z2 = 0
(1)
4A
4(1.7h2 )
=
= 1.26h
Also, Dh =
P 2(1.7h + h)
h
b = 1.7h
and
m3
0.006
Q
s = 0.00353h −2 m , where h is in m
V= =
2
A
s
1.7 h
Thus, from Eq. (1),

 7 m    3.53 ×10
4.07 m = 1 + f 

h2
 1.26h m   

or


m
1

 2 
 s  2  9.81 m  
 
s2  
−3 2

2 
f
h
From Table 8.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7) and
Colebrook (Ref. 8)],
(2) 6.41×106 h 4 = 1 + 5.55
ε
Dh
=
0.15 ×10 −3 m 1.19 ×10 −4
=
, where h ~ m (3)
h
1.26h m
and
m

0.00353h −2  (1.26 h m)

VDh 
3970
s 
=
Re h =
or Re h =
2
v
h
m
1.12 × 10 −6
s
(4)
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
103
4
6
2(104)
8
104
4
6
2(105)
8
4
6
105
2(106)
8
106
4
6
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Figure (1)
Finally, from Figure 1 above:
Trial and error solution of Eqs. (2), (3), (4), and Figure 1 for f, h, Reh,
Assume h = 0.04 m; from (2) f = 0.111, from (3),
104. Hence, from the figure f = 0.0223 ≠ 0.111
ε
Dh
ε
Dh
.
= 1.07 ×10 −3, and from (4), Reh = 9.93 ×
Assume h = 0.03 m; from (2) f = 0.0227,
the figure f = 0.0290 ≠ 0.0227
Assume h = 0.025 m; or f = 0.00677,
figure f = 0.0303 ≠ 0.00677
Assume h = 0.035 m; or f = 0.0544,
ε
Dh
ε
Dh
ε
Dh
= 4.0 ×10 −3, and Reh = 1.32 × 105. Hence, from
= 4.76 ×10−3, and Reh = 1.59 × 105. Hence, from the
= 3.40 ×10−3, Reh = 1.13 × 105. Hence, from the
figure f = 0.0280
Plot f from Eq. (2) and f from Figure 1 as a function of h. Solution is where the two
curves intersect.
0.06
•
From Eq. (5)
0.04
f
0.02
From Eq. (2)
•
0
0.025
h, m
0.035
Thus, h = 0.031 m and b = 1.7 (0.031 m)
or 0.031 m by 0.053 m
This problem can be solved using the Colebrook equation, rather than the Moody chart,
Figure 1 above, as follows:
 ε

 D
1
2.51 
= −2.0 log 
+
From

3.7
f
Re f 



(6)
where from Eq. (2),
(6.41×106 h5 − h )
(7)
5.55
Combining Eqs. (3), (4), (6), and (7) gives a single equation for h:
f =




1
2.51h
1.19 ×10 −4

= −2.0 log 
+

6
5
6
5
 (6.41×10 h − h ) 
 (6.41×10 h − h )  
 3.7 h
3970 




5.55
5.55

 



(8)
Using a computer root-finding program gives h = 0.0313 m, which is the same as that
obtained by the above trial and error method.
Problem 8.107
Rainwater flows through the galvanized iron downspout shown in the figure below at a rate
m3
of 0.006
. Determine the size of the downspout diameter if it is circular and it is
s
completely filled with water. Neglect the velocity of the water in the gutter at the free
surface and the head loss associated with the elbow.
70 mm
g
4m
3m
Solution 8.107
p1 V12
p2 V22
 V2
+
+z =
+
+z + f
, where p1 = p2 = 0, V1 = 0, V2 = V1, z1 = 4.07 m, and
γ 2g 1 γ 2g 2
Dh 2 g
2
 V

z2 = 0 Thus, z1 = 1 + f 
or
D  2g

m 

 7 m  2
(4.07 m)(2)  9.81 2  = 1 + f 
 V
 D 
s  

Hence, with V =
Q
π
4
D
or
2
 7f   0.00764 
79.9 = 1 +  

D   D2 

(1)

m3 
4  0.006


s  0.00764

, Eq. (1) becomes
=
V=
π D2
D2
2
or f = 1.956 × 105 D5 - 0.1429 D, where D ~ m
(2)
 0.00764 
m
D
0.00764


2
VD  D
 =
s
Also, Re =
=
2
v
v

−6 m 
×
1.12
10

 D
s


or
Re =
6.82 ×103
D
(3)
From Table 8.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7) and
Colebrook (Ref. 8)]
0.15 ×10−3
(4)
D
D
so that the Colebrook equation,
ε
=
 ε

 D
1
2.51 
= −2.0 log 
+
 can be combined with Eqs. (3) and (4).
3.7
f
Re
f




 4.05 ×10 −5 3.68 ×10 −4 D 
1
= −2.0 log 
+

D
f
f


(5)
Solve Eqs. (2) and (5) for f and D as follows: substitute f from Eq. (2) into Eq. (5) to
obtain a single equation for D:


 4.05 × 10 −5

3.68 × 10 −4 D
= −2.0 log 
+

1
1
D

5 5
5 5
2
2
(1.956 × 10 D − 0.1429D )
(1.956 × 10 D − 0.1429D ) 

1
Using a computer root-finding technique gives
D = 0.0445 m
(6)
Problem 8.108
For a given head loss per unit length, what effect on the flowrate does doubling the pipe
diameter have if the flow is (a) laminar, or (b) completely turbulent?
Solution 8.108
(a) hL = f
hL =
 V2
ε 
64

,
, where f = f  Re,  . Thus, for laminar flow with f =
D
D 2g
Re

γ D2 hL
64ν  V 2 32 μ V
=
V
or
=
32μl
VD D 2 g
γ D2
Thus, since Q =
Q=
π
4
D2V , it follows that
πγ D hL
If hL, γ, μ, and ℓ are fixed, doubling D will increase Q by a factor of 24
128μl
4
= 16 for laminar flow.
(b) For completely turbulent flow, f is a function of
ε
D
only (not Re).
Thus,
1
1
 2 gDhL  2
π 2  2 gDhL  2
V =
 or with hL, g, and ℓ constant
 or Q = D 
4
 f 
 f 
Q~
5
D2
(1)
f
ε 
From the figure below, for very large Re flow f = f   and f increases with
D
increasing
ε
D
, although the dependence is not strong. If we neglect this dependence
5
and assume f is constant when the diameter is doubled, then Q ~ D 2 , so that the
5
flowrate would increase by a factor of 2 2 = 5.66. If we account for the
dependence we obtain from Eq. (1)
ε 
f 
D
0.1
0.09
Wholly turbulent flow
0.08
Laminar flow
0.05
0.07
Transition range
0.04
0.06
0.03
0.05
0.02
0.015
0.04
0.01
0.008
0.006
0.03
0.004
f
ε
__
D
0.025
0.002
0.02
0.001
0.0008
0.0006
0.0004
0.015
0.0002
Smooth
0.0001
0.00005
0.01
0.009
0.008
2(103)
4
6
103
2(104)
8
4
104
6
2(105)
8
4
105
6
2(106)
8
4
6
106
2(107)
8
4
6
8
0.00001
107
ρ VD
Re = _____
μ
Q2D
=
QD
and
ε
5
(2D ) 2
fD
5
(D ) 2
f2 D
2D
Thus,
5
= 22
fD
ε
, where typical values for fD and f2D (i.e., with = 0.002
f2D
D
= 0.001) for large Re are fD = 0.0232 and f2D = 0.0195
5
Q2 D
0.0232
= 22
= 6.17
QD
0.0195
Note: If we consider
ε
D
= 0.02 and
ε
2D
= 0.01, or f D = 0.0483 and f 2D = 0.0382 we
obtain
5
Q2 D
0.0483
= 22
= 6.36
QD
0.0382
Thus, for large Re turbulent flow doubling the diameter increases the flowrate by a
factor of about 6.36.
Problem 8.109
It is necessary to deliver 270 ft3/min of water from reservoir A to reservoir B, as shown in
the figure below. The connecting piping consists of four fully open gate valves, 12 regular
90° elbows, one swing check valve, two fully open globe valves, and 3000 ft of commercial
steel pipe. Find the minimum diameter for steel pipe needed to obtain this flowrate. All
fittings and valves have flanged connections. Assume a square-edged entrance.
Elevation 150.0 ft
Reservoir A
Elevation 25.0 ft
Reservoir B
Solution 8.109
Write the mechanical energy equation from the reservoir A water surface (a) to the reservoir
B water surface (b).
pa Va2
pb Vb2
+
+ gza + ghp =
+
+ gzb + ght + ghL
ρ
2
ρ
2
Now pa = pb and Va ≈ Vb ≈ 0. Also ght = ghp = 0, so we get
ghL = g(za − zb)
where
2
 
V
ghL =  f + K L ent + 4 K L gate + 12 K L 90° + K L check + 2 K L globe + K L exit 
 D
 2
2
 
 8Q
=  f + K L ent + 4 K L gate + 12 K L 90° + K L check + 2 K L globe + K L exit  2 4
 D
π D
The energy equation is
2
 
 8Q
 f + K L ent + 4K L gate + 12K L 90° + K L check + 2K L globe + K L exit  2 4 = g( za − zb )
 D
π D
or
2
1  
 π g ( za − zb )
+
+
+
+
+
=
f
K
4
K
12
K
2
K
K
° + K
L ent
L gate
L check
L globe
L exit 

L 90
8Q 2

D4  D
For a square-edged entrance, standard elbows and a swing check valve, Table 8.2 and
Figure 8.22 give
KL ent = 0.5, KL check = 2.0, and KL exit = 1.0.
Also
π 2 g ( za − zb )
8Q
2
=
π 2 (32.2 ft / s2 )(125 ft)
3
2  min 
2
=
8(270 ft / min) 

 60 s 
245
ft 4
and the energy equation gives
1  f (3000 ft)
 245
+ 0.5 + 4K L gate + 12K L 90o + 2.0 + 2K L globe + 1.0  = 4
4 
D
D 
 ft
or
1 
f
 245
(3000 ft) + 3.5 + 4K L gate + 12K L 90o + 2K L globe  = 4
4 
D
D 
 ft
(1)
Assuming 60°F water,
 min 
4(270 ft3 / min) 
5

VD 4Q
 60 s  = 4.74 ×10
Re =
=
=
v
D
π Dv π (1.21×10 −5 ft 2 / s)D
(2)
and ε = 0.00015 ft. Equations (1) and (2) and the Moody chart must be solved for D by
trial and error. Guess D = 0.75 ft. Then
ε
D
=
(0.00015 ft)
= 0.00020,
(0.75 ft)
(4.74 ×105 ft)
Re =
= 6.32 ×105 ,
(0.75 ft)
the Moody chart gives f = 0.015, and Table 8.2 gives
KL gate = 0.15, KL 90° = 0.3, and KL globe = 10.
The left-hand side of Eq. (1) gives
 277
1  (3000 ft)
(0.015) + 3.5 + 4(0.15) + 12(0.3) + 2(10)  = 4

(0.75)  (0.75)
 ft
For D = 0.77 ft, the solution converges so
D = 0.77 ft.
Problem 8.110
A 10-m-long, 5.042-cm I.D. copper pipe has two fully open gate valves, a swing check
valve, and a sudden enlargement to a 9.919-cm I.D. copper pipe. The 9.919-cm copper pipe
is 5.0 m long and then has a sudden contraction to another 5.042-cm copper pipe. Find the
head loss for a 20 °C water flowrate of 0.05 m3/s.
Solution 8.110
1
gate
swing
check
gate
2
3
ID = 5.042 cm
ID = 9.919 cm
ℓA = 10 m
ℓB = 5 m
Q = 0.05 m3/s
20 °C water
The energy loss is
ghL = ( ghL )1−2 + ( ghL )2 −3
where the loss coefficients for the enlargement and contraction are based on the velocity in
the 5.042 cm pipe. This gives

 V 2   V 2
ghL =  2K L gate + K L check + K L enl + K L cont + f A  A +  f B  B
DA  2  DB  2



  8Q 2
  8Q 2
=  2K L gate + K L check + K L enl + K L cont + fA A  2 4 +  fB B  2 4 .
DA  π DA  DB  π DB

Table 8.2, Figure 8.26 (or the equation), and Figure 8.25 give
K L gate = 0.15, K L check = 2
2
2
  D 2 
  5.042 cm 2 
A
K L enl 1 − 
  = 1 − 
  = 0.55,
  DB  
  9.919 cm  


and
K L cont ≈ 0.4
Using Table B.2, the Reynolds numbers are
 VD 
 4Q 
Re A = 
 =
 =
 v A  π Dv A
4(0.05m3 / s)
= 1.3 ×106
2

m
 5.042 
π
m  1.0 ×10 −6


s 
 100

and
 VD 
 4Q 
Re B = 
 =
 =
 v B  π Dv B
4(0.05m3 / s)
= 6.4 ×105
2
m 
 9.919  
π
m  1.0 ×10 −6

s 
 100
 
Using Table 8.1 for copper pipe (drawn tubing),
0.0015 mm
ε 
= 0.000030
  =
 D A 50.42 mm
and
0.0015 mm
ε 
= 0.000015.
  =
 D B 99.19 mm
The Moody chart gives
fA = 0.0118 and fB = 0.013.
The energy loss is
2
3

 8  0.05 m 

s 
0.0118(10 m)  

ghL =  2 ( 0.15 ) + 2 + 0.55 + 0.4 +
 5.042   2  5.042 4

m π

m

 100
   100


2
3

 8  0.05 m 
 0.013 ( 5 m )  
s 
m2  N ⋅ s2 
 
+
=
1767


  9.919 m   2  9.919 4
s2  kg ⋅ m 
 π 
m
  100
   100


ghL = 1767
N⋅m
kg
Problem 8.111
Air, assumed incompressible, flows through the two pipes shown in the figure below.
Determine the flowrate if minor losses are neglected and the friction factor in each pipe is
0.015. Determine the flowrate if the 0.5-in.-diameter pipe were replaced by a 1-in.-diameter
pipe. Comment on the assumption of incompressibility.
p = 0.5 psi
T = 150°F
1 in.
0.50 in.
20 ft
20 ft
Solution 8.111
p = 0.5 psi
T = 150°F
1 in.
0.50 in.
(1)
(2)
20 ft
20 ft
(0)
(3)
•
p0 V02
p V2
+
+ z0 = hL1 + hL2 + 3 + 3 + z3 , where V0 = 0, z0 = z2 , p3 = 0,V2 = V3 , (1)
γ 2g
γ 2g
2
2
2
 D2   0.5 in. 
 1 V1
 2 V22
A2
hL1 = f1
, hL2 = f2
, and V1 = V2
= V2 
 =
 V2 = 0.25V2
D1 2 g
D2 2 g
A1
 D1   1.0 in. 
Thus, Eq. (1) becomes
p0
γ
= f1
 1 (0.25V2 )2
 V2 V2
+ f2 2 2 + 2
D1
2g
D2 2 g 2 g
or
p0 =
 


1
ρV22  f1 1 (0.25)2 + f2 2 + 1
D2 
2
 D1
lb
lb  
in.2 

+
0.5
14.7
144




in.2
in.2  
ft 2 

p0
slug
With p0 = ρ0RT0 or ρ0 =
=
= 0.00209 3
RT0

ft ⋅ lb 
ft
 1716 slug ⋅ °R  (150 + 460)°R


and f1 = f2 = 0.015. Equation (2) gives




 
  20 ft 

 
lb  
in.2  1 
slug  2 
20
ft

2
 0.5 2  144 2  = 2  0.00209 3 V2 (0.015)   1  (0.25) +  1   + 1
in.  
ft 
ft  


  ft 
 ft   

 24   
  12 
(2)
or
2
π 1  
ft 
ft
ft
ft   90.4  = 0.123
V2 = 90.4 . Thus, Q = A2V2 = 
4  24  
s
s
s
3
If both pipes were 1 in. diameter, then V1 = V2 and Eq. (1) becomes
p0 =
 


1
ρV22  f1 1 + f 2 + 1 or with f1 = f2, ℓ1 = ℓ2, D1 = D2
D2 
2
 D1
p0 =
 2

1
ρV22  f2 2 + 1
2
 D2

Hence,


 
2


 40 ft  
lb 
in.
1
slug  2

=
0.5
144
0.00209
0.015
V



 1  + 1



 2
in.2  
ft 2  2 
ft3  

 ft  

 12  
or
2
V2 = 91.7
π1  
ft 
ft
ft
Thus, Q = A2V2 =  ft   91.7  = 0.500
4  12  
s
s
s
3
Since p = ρRT it follows that
 p3 


ρ3  RT3  p3 T0
If we assume T3 = T0 (it probably will not be but it should be a
=
=
ρ0  p0  p0 T3


 RT0 
reasonable approximation) then
ρ3 p3
14.7 psi
≈
=
= 0.967
ρ0 p0 (0.5 + 14.7) psi
The flow is nearly incompressible.
Problem 8.112
Air, assumed incompressible, flows through the two pipes shown in the figure below.
Determine the flowrate if minor losses are neglected, the pipes are galvanized iron and the
friction fsactors are not known a priori. Determine the flowrate if the 0.5-in.-diameter pipe
were replaced by a 1-in.- diameter pipe. Comment on the assumption of incompressibility.
p = 0.5 psi
T = 150°F
1 in.
0.50 in.
20 ft
20 ft
Solution 8.112
p = 0.5 psi
T = 150°F
1 in.
0.50 in.
(1)
(2)
20 ft
20 ft
(0)
(3)
•
p0 V02
p V2
+
+ z0 = hL1 + hL2 + 3 + 3 + z3 , where V0 = 0, z0 = z2 , p3 = 0,V2 = V3 ,
γ 2g
γ 2g
2
(1)
2
 D   0.5 in. 
 V2
 V2
A
hL1 = f1 1 1 , hL2 = f2 2 2 , and V1 = V2 2 = V2  2  = 
 V2 = 0.25V2
D1 2 g
D2 2 g
A1
 D1   1.0 in. 
Thus, Eq. (1) becomes
p0
γ
= f1
 1 (0.25V2 )2
 V2 V2
+ f2 2 2 + 2
D1
2g
D2 2 g 2 g
or
p0 =
 


1
ρV22  f1 1 (0.25)2 + f2 2 + 1
D2 
2
 D1
(2)
lb
lb  
in.2 

+
0.5
14.7
144




in.2
in.2  
ft 2 

p0
slug
With ρ0 = ρ0RT0 or ρ0 =
=
= 0.00209 3 .
RT0

ft ⋅ lb 
ft
 1716 slug ⋅°R  (150 + 460)°R


Equation (2) becomes




 
2





lb 
in.
1
slug  2
20 ft
20 ft  

2
0.5
144
0.00209
V
(0.25)
f
f
=
+
+1






1
2



 2
1
1  
in.2  
ft 2  2 
ft3  

 ft 
 ft  
 12 
 24  

or
(3)
6.89 ×104 = V22 (15 f1 + 480 f2 + 1)
(3)
Also from Table 8.1 Equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7)
and Colebrook (Ref. 8)]
ε
D1
=
0.0005 ft 0.0005 ft
=
= 0.006
1
D1
ft
12
(4)
and
ε
D2
=
0.0005 ft
= 0.012
1
ft
24
(5)
and
V1D1
VD
,Re2 = 2 2 , where from Table B.2 Physical Properties of Air at Standard
v
v
Atmospheric Pressure (BG/EE Units)
Re1 =
4.18 ×10 −7
lb ⋅ s
2
μ
ft 2 = 2.00 ×10 −4 ft
v=
=
slug
s
ρ0
0.00209 3
ft
Hence,
 1 
(0.25V2 )  ft 
 12  = 104V
Re1 =
2
2
−4 ft
2.00 × 10
s
(6)
and
 1 
V2  ft 
 24  = 208V
Re2 =
2
2
−4 ft
2.00 × 10
s
(7)
 ε


1
2.51 
= −2.0 log  D +
For turbulent flow, the Colebrook equation gives

f
 3.7 Re f 


(8)
By combining Eqs. (4) through (8), we obtain

1
2.41×10 −2 
= −2.0 log 1.62 ×10 −3 +

f1
V2 f1 

(9)
and

1
1.21×10 −2 
= −2.0 log 3.24 ×10−3 +

f2
V2 f2 

(10)
A computer trial and error solution method gives the solution to Eqs. (3), (9), and (10) as:
f1 = 0.0425, f2 = 0.0446 , and V2 = 54.7
ft
s
Thus,
Q = A2V2 =
π
4
π  0.50 
D22V2 = 
ft 
2
3
ft 

−2 ft
 54.7  = 7.46 ×10
s
s
 
4  12
If D1 = D2 , then V1 = V2, f1 = f2 since
ε
D1
=
ε
D2
= 0.006 , and
 1

V2 
ft 
VD
 12  = 416V
Re1 = Re 2 = 2 2 =
2
2
v
−4 ft
2.00 × 10
s
Thus, Eq. (1) becomes
p0 =
  +  
1
ρV22  f2  1 2  + 1
2
  D2  
or
 
 
2



 
lb
in.
1
slug
40
ft



 2
 0.5 2  144 2  = 2  0.00209 3 V2  f2  1  + 1
in.  
ft 
ft    ft  


  12  
Hence,
6.89 ×104 = V22 [480 f2 + 1]
(11)
Also, from Eq. (8),

1
6.03 ×10 −2 
= −2.0 log 1.62 ×10 −3 +

f2
V2 f2 

(12)
A computer solution of Eqs. (11) and (12) gives
f2 = 0.0351 and V2 = 62.2
ft
s
Thus,
Q = A2V2 =
π
4
D22V2 =
π1
2
ft 
ft3
 
ft
62.2
0.339
=

 

4  12  
s
s
Note: Since p = ρRT, it follows that
 p3 


ρ3  RT3  p3 T0
, if we assume T3 = T0 (it probably will not be, but it should be a
=
=
ρ0  p0  p0 T3


 RT0 
reasonable approximation) then
ρ3 p3
14.7 psi
=
=
= 0.967
ρ0 p0 (0.5 + 14.7) psi
The flow is nearly incompressible.
Problem 8.114
Normal octane at 68 °F (ν = 8.31 × 10−6 ft2/s) is to be delivered at a flowrate of 3.0 gal/min
through a 2.0-in. schedule 40 commercial steel pipe (with an actual inside diameter of 2.067
in.). Energy losses are important, so consideration is being given to expanding the pipe to a
3.0 in. schedule 40 commercial steel pipe (with an actual inside diameter of 3.068 in.), using
a 10° conical expansion. After a length ℓ of the 3.0-in. pipe, it is decreased to a 2.0-in.
schedule 40 commercial steel pipe with a gradual contraction (K L = 0.10 , based on the
smaller diameter pipe flow). Find the minimum length ℓ so that the energy loss will be the
same as in a 2.0-in. pipe of length ℓ + 2(5.7 in.). See the figure below.
2-in. pipe
5.7 in.
5.7 in.
𝓵
Expansion
3-in. pipe
Contraction
Solution 8.114
Consider the two piping systems:
D2
5.7 in.
5.7 in.
L
D2
D2
D3
The energy loss in the 2.0 in. pipe system is
ghL = f2
( L + 11.4 in.) V2 2
8Q 2
= f2 ( L + 11.4in.) 2 5 .
D
2
π D2
The energy loss in the 3.0 in. pipe system is
(
ghL = K L exp + K L cont
)
V22
L V32
8Q 2
8Q 2
+ f3
= K L exp + K L cont 2 4 + f3 L 2 5 .
D3 2
2
π D2
π D3
(
)
The problem statement gives KL cont = 0.10 and Figure 8.28 gives KL exp ≈ 0.4. Equating the
two equations for ghL gives,
f2 ( L + 11.4 in.)
f2
8Q 2
π 2D2
( L + 11.4 in.)
D25
=
(
= K L exp + K L cont
5
( K L exp + K L cont )
D2
4
+
2
2
2
3
) π82QD 5 + f3 π82QD 5 ,
f3L
D35
or
 K L exp + K L cont 11.4 in. f2 
−


D2 4
D25 

L=
.
 f2
f3 
 5 − 5 
 D2 D3 
With the given diameters, D2 = 2.067 in. and D3 = 3.068 in., and known numerical values,
 0.4 + 0.1
(11.4 in.) f2 
−


(2.067 in.)4 (3.068 in.)5 

L=
.


f3
f2
−

5
5

 (2.067 in.) (2.068 in.) 
L=
(0.5 − 0.766 f2 )in.
(1 − 1.532 f2 )in.
.
=
(0.484 f2 − 0.139 f3 ) (0.968 f2 − 0.278 f3 )
The friction factors are found using the Moody chart.
Re2 =
ε2
D2
=
D3
=
ν
=
4Q
=
π Dν
ft 2  
s 
 2.067  
ft   8.31×10−6
π
  60


s   min 
 12

VD
ν
=
4Q
=
π Dν
gal   ft3 

4  3.0


min   7.48 gal 

s 
ft 2  
 3.068  
ft   8.31×10 −6
60
π



s   min 
 12
 
0.00015 ft
= 0.00059, and f3 = 0.041
 3.068 
ft 

 12

The length L is
[1 − 1.532(0.037)] in.
[0.968(0.037) − 0.278(0.041)]
L = 38.6 in.
L=
= 5950,
0.00015 ft
= 0.00087, f2 = 0.037,
 2.067 
ft 

 12

Re3 =
ε3
VD
gal   ft3 

4  3.0


min   7.48 gal 

= 4010,
Problem 8.115
The flowrate between tank A and tank B shown in the figure below is to be increased by
30% (i.e., from Q to 1.30 Q ) by the addition of a second pipe (indicated by the dotted lines)
running from node C to tank B. If the elevation of the free surface in tank A is 25 ft above
that in tank B, determine the diameter, D, of this new pipe. Neglect minor losses and
assume that the friction factor for each pipe is 0.02.
6-in. diameter;
600 ft long
6-in. diameter;
500 ft long
C
A
B
Diameter D, 500 ft long
Solution 8.115
(A)
•
6-in. diameter;
600 ft long
6-in. diameter;
500 ft long
(B)
•
C
A
(1)
(2)
B
(3)
Diameter D, 500 ft long
With the single pipe:
pA VA2
p
V2
 V2
 V2
+
+ zA = B + B + zB + f1 1 1 + f2 2 2
γ
2g
γ
2g
D1 2 g
D2 2 g
where pA = pB = 0, VA = VB = 0, zA = 25 ft, zB = 0, and V1 = V2 (since D1 = D2).

Thus, zA = f1

 (600 + 500)ft 
(  1 +  2 ) V12

, or 25 ft = (0.02) 
D1



2g
6 
  ft
 12 
 2  32.2 ft 

 
s2 
 
or
2
π6  
ft 
ft 3
ft
V1 = 6.05 . Hence, Q = AV
1 1=
 ft   6.05  = 1.188
4  12  
s
s
s

ft 3 
ft 3
With the second pipe, Q = (1.30)  1.188
= 1.54


s 
s

ft3
ft
Q
s = 7.84 ft
= Q2 + Q3 or V1 = 1 =
Thus, Q1 = 1.54
s
A1 π  6 2
s
 ft 
4  12 
3
1.54
V12
(1)
For fluid flowing from A to B through pipes 1 and 2,
 1 V12
 2 V22
[see Eq. (1)]
zA = hL1 + hL2 = f1
+ f2
D1 2 g
D2 2 g
or
2
ft 

600 ft  7.84 
V22
500 ft
s

25 ft = (0.02)
+ (0.02)
ft 
ft 
6  
6  
 ft  2  32.2 2 
 ft  2  32.2 2 
 12  
s 
s 
 12  
Hence, V2 = 2.60
and Q2 = A2V2 =
ft
s
π6
2
ft 
ft 3
 
ft
2.60
0.511
=

 

4  12  
s
s
ft3
ft3
ft3
− 0.511
= 1.03
Thus, Q3 = Q1 − Q2 = 1.54
s
s
s
For fluid flowing from A to B through pipes 1 and 3,
ft3
1.03
Q
 V2
 V2
s
zA = hL1 + hL3 = f1 1 1 + f3 3 3 where V3 = 3 =
π
A3
D1 2 g
D3 2 g
D32
4
=
1.31
D32
Thus,
2
2
 1.31 
ft 

 2 
7.84

600 ft 
500 ft  D3 
s
25 ft = (0.02)
+ (0.02)
ft 
ft 
D3
6  

2  32.2 2 
 ft  2  32.2 2 
 12  
s 
s 

or
D2 = 0.662 ft
Note: With the parameters given, the solution is quite sensitive to round off errors in the
calculations.
Problem 8.116
A 250-ft-high building has a 6.065-in.-diameter steel standpipe and a 100-ft-long
9
2 -in.-diameter fire hose on each floor. The nearest fireplug is 100 ft from the standpipe’s
16
ground-level connection. Assume that fire-fighters connect a 6-in.-diameter, 50-ft-long fire
hose from the fireplug to the fire truck and a 4-in.-diameter, 50-ft-long fire hose from the
fire truck to the standpipe’s ground-level connection. The National Fire Protection
Association (NFPA) requires that a minimum pressure of 65 psig be maintained at the
9
connection of the 2 -in.-diameter hose and the standpipe while maintaining a flowrate of
16
500 gal/min through the fire hose. What pressure rise must the pump on the fire engine
supply to satisfy the NFPA requirement for this building? The fire hydrant water pressure
is 80 psig and the water temperature is 60 °F. The connections are threaded.
Solution 8.116
Pmin = 65 psig
(2)
Q = 500 gpm
9
Twater = 60 °F
L = 250'
__
Dd = 216
″ firehose
• c
b
AERIAL LADDER
Dc = 6″ standpipe
Db = 4″ firehose
ℓb = 50'
Fireplug
a
Da = 6″ firehose
ℓa = 50'
(1)
p = 80 psig
Assume steady state and apply the mechanical energy equation to a streamline from point 1
to point 2 with ht = 0 to get
p 2 V22
p1 α1V12
p α V2
p V2
+
+ gz1 + ghp = 2 + 2 2 + gz2 + ghL 1 + 1 + gz1 + ghp =
+
+ gz2 + ghL
ρ
ρ
ρ
ρ
2
2
2
2
Solving for hp gives
hp =
p2 − p1
γ
+ ( z2 − z1 ) +
2
V22 V12
 f
V
−
+ 
+ KL 
2g 2g
D
 2g
To evaluate f for each hose and the standpipe, assume the hose is equivalent to galvanized
iron so Table 8.1 gives εa = εb = 0.00085 ft and εc = 0.00015 ft. Assume the firehose diameter
is the inside diameter and the stand pipe diameter is given as Dc = 6.065 in. assuming a
schedule 40 pipe. Then
εa
Da
=
ε
0.00085 ft
0.00085 ft
= 0.0017, b =
= 0.0025,
Db
6 
4 
 ft 
 ft 
 12 
 12 
and
εc
Dc
=
0.00015 ft
= 0.00030.
 6.065 
ft 

 12

Table B.1 gives ν = 1.21×10−5 ft 2 /s for 60 °F water. So
 π D2 
V
 4  4
VD
4Q

= 
=
Re =
,
ν
ν
π D π Dν


ft3 / s
4(500 gal / min) 
 448.8 gal / min 

 = 2.34 × 105 ,
Rea =
6 
π  ft  (1.21×10 −5 ft 2 / s)
 12 


ft3 / s
4(500 gal / min) 
 448.8 gal / min 

 = 3.52 ×105 ,
Reb =
 4 
π  ft  (1.21×10 −5 ft 2 / s)
 12 
and


ft3 / s
4(500 gal / min) 
 448.8 gal / min 

 = 2.32 × 105 ,
Rec =
 6.065 
ft  (1.21×10 −5 ft 2 / s)
π
12


The Moody chart gives
fa = 0.0233, fb = 0.0253, and fc = 0.0174
Assume the loss coefficient at the 90° at the standpipe inlet and outlet can be modeled by
90° standard, threaded elbows (KL = 1.5). The numerical values give
2
2
2
1
 
V
 
  4Q  16Q
 
 1
  f D + K L  2 g = 2 g   f D + K L   π D2  = 2π 2 g   f D + K L  D4
4

4

 

 


 

0.0233 (50 ft)  1  +  0.0253 (50 ft) + 1.5  1  



 6   6 ft  
 4 
  4   
ft


  12  
 ft 
  12 ft   
12 
12 

 





16(500 gal / min)2

 = 41.2 ft
4
=


2 

4
 32.2 ft  448.8 gal / min  





π 2  2 

3





(250 ft)
1
1
ft / s
 s

 
 + 1.5 

 +0.0174 6.065  6.065

9 









2
ft
ft



 16 


 12
   12

ft



 12 

4
4
V22 −V12 8Q 2  1   1  
= 2 
 −
 
2g
π g  Dd   Da  
4



4





  1  
8(500 gal / min)2
1

 
=

 −
2 
9
6
   ft   
 32.2 ft   448.8 gal / min   2

π 2  2 
  16 ft    12   
ft3 / s
 s

 


 12 

= 14.5 ft,
and
lb 
in.2 
144


in.2 
ft 2 
+ 250 ft + 41.2 ft + 14.5 ft = 271 ft
lb 

 62.4 3 
ft 

(65 − 80)
hp =
The pump pressure rise, assuming sea level elevation, is
 ft 2 
lb 

 p = ρ ghp =  62.4 3  (271ft) 
 144 in.2 
ft 



or
p = 117 psi.
Problem 8.117
With the valve closed, water flows from tank A to tank B as shown in the figure below.
What is the flowrate into tank B when the valve is opened to allow water to flow into tank
C also? Neglect all minor losses and assume that the friction factor is 0.02 for all pipes.
Elevation = 15 m
Diameter of each pipe = 0.10 m
A
Elevations = 0
80 m
40 m
B
C
75 m
Solution 8.117
Elevation = 15 m
• (A)
Diameter of each pipe = 0.10 m
A
Elevations = 0
80 m
40 m
(1)
(2)
• (B) (C) •
B
C
(3)
75 m
Q1 = Q2 + Q3 where Qi = AV
i i =
π
4
Di2Vi, i = 1, 2, 3
Thus, since D1 = D2 = D3, it follows that
V1 = V2 + V3
(1)
Also, for fluid flowing from A to B,
pA VA2
pB VB2
 1 V12
 2 V22
+
+ zA =
+
+ zB + f1
+ f2
, where pA = pB = 0, VA = VB = 0, zA = 15
γ
2g
γ
2g
D1 2 g
D2 2 g
m, and zB = 0.
Thus, zA = f1
15 m =
 1 V12
 V2
+ f2 2 2 ,
D1 2 g
D2 2 g
or
(2)
(0.02)
[(80 m)V12 + (40 m)V22 ]
m

(0.1 m)(2)  9.81 2 
s 

Hence,
18.4 = V12 + 0.5V22 , Vi ∼
m
s
(3)
Similarly, for fluid flowing from A to C,
p V2
 V2
pA VA2
 V2
+
+ zA = C + C + zC + f1 1 1 + f3 3 3 , where pA = pC = 0, VA = VC = 0, zA = 15
γ 2g
γ
2g
D1 2 g
D3 2 g
m, and zC = 0.
Thus, zA = f1
 V2
 1 V12
+ f3 3 3 , or
D1 2 g
D3 2 g
By comparing Eqs. (2) and (4), we find f2
(4)
 V2
 2 V22
= f3 3 3
D2 2 g
D3 2 g
or since f2 = f3 and D2 = D3,
=
Thus,
=
or V2 = 1.369V3
(5)
Solve Eqs. (1), (3), and (5) for V1, V2, and V3.
From Eqs. (1) and (5): V2 = 1.369V3 + V3 = 2.369V and from Eq. (3),
18.4 = (2.369V3)2 + 0.5(1.369V3)2 or V3 = 1.676
V1 = 2.29
m
m
m
+ 1.676 = 4.0
s
s
s
Thus
Q2 = A2V2 =
π
m
m3

(0.1 m)2  2.29  = 0.0180
4
s 
s

m
m

m
, V2 = 1.369 1.676  = 2.29 , and
s 
s
s

Problem 8.118
With the valve closed, water flows from tank A to tank B as shown in the figure below.
What is the flowrate into tank B when the valve is opened to allow water to flow into tank
C also? Neglect all minor losses and assume the friction factors are not known, but the
pipes are steel pipes.
Elevation = 15 m
Diameter of each pipe = 0.10 m
A
Elevations = 0
80 m
40 m
B
C
75 m
Solution 8.118
Elevation = 15 m
• (A)
Diameter of each pipe = 0.10 m
A
Elevations = 0
80 m
40 m
(1)
(2)
• (B) (C) •
B
C
(3)
75 m
Q1 = Q2 + Q3 where Qi = AV
i i =
π
4
Di2Vi , i = 1, 2, 3
Thus, since D1 = D2 = D3 it follows that
V1 = V2 + V3
(1)
Also, for fluid flowing from A to B,
 V2
 V2
pA VA2
p
V2
+
+ zA = B + B + zB + f1 1 1 + f2 2 2 , where pA = pB = 0, VA = VB = 0, zA = 15
γ
2g
γ
2g
D1 2 g
D2 2 g
m, and zB = 0.
 1 V12
 2 V22
Thus, zA = f1
+ f2
, or
D1 2 g
D2 2 g
(2)
  80 m  2
 40 m  2 
1
15 m =  f1 
V1 + f2 
V2 
 0.1 m   2  9.81 m 
  0.1 m 


s2 

or
0.368 = f1V12 + 0.5 f2V22
(3)
Similarly, for fluid flowing from A to C,
p V2
 V2
pA VA2
 V2
+
+ zA = C + C + zC + f1 1 1 + f3 3 3 , where pA = pC = 0, VA = VC = 0, zA =
γ 2g
γ
2g
D1 2 g
D3 2 g
15 m, and zC = 0.
Thus, zA = f1
 V2
 1 V12
+ f3 3 3 , or
D1 2 g
D3 2 g
By comparing Eqs. (2) and (4), we find f2
(4)
 V2
 2 V22
= f3 3 3
D2 2 g
D3 2 g
or since and D2 = D3,
1
 f3  2
f2  2V22 = f3  3V32 Thus, 40 f2V22 = 75 f3V32 or V2 = 1.369   V3
 f2 
(5)
 ε

 D
1
2.51 
= −2.0 log 
+
From the Colebrook equation,
 where from Table 8.1
3.7
f
Re
f




equivalent Roughness for New Pipes [Adapted from Moody (Ref. 7) and
0.045 mm
Colebrook (Ref. 8)] ε = 0.045 mm so that for each pipe, ε =
= 4.5 × 10 −4
100 mm
Also, Re =
VD
or for i = 1, 2, 3, and using Table 1.6
v
Rei =
Vi Di
=
v
Vi (0.1 m)
Thus,

1
2.81× 10 −5 
= −2.0 log 1.22 × 10 −4 +
 for i = 1, 2, 3
fi
Vi fi 

1.12 ×10 −6
m
s
2
= 8.93 × 10 −4Vi
(6), (7), (8)
Solve six equations for six unknowns: Eqs. (1), (3), (5), (6), (7), and (8) for f1, f2, f3, V1, V2,
V3. Trial and error solution as follows:
1
 f 2
From Eq. (5), V3 = 0.730  2  V2, which when combined with Eq. (1) gives
 f3 
1

 f 2 

V1 = 1 + 0.730  2  V2
 f3  



Thus, by combining Eqs. (3) and (9), we obtain
 

1 2
 

 f 2 
0.368 =  f1 1 + 0.730  2   + 0.5 f2 V22
 f3  
 





(9)
or
1

2






0.368


V2 = 

2
1
 

 f2  2 
 

 f1 1 + 0.730  f   + 0.5 f2 
 3 
 




(10)
Also, from Eq. (1),
V3 = V1 − V2
(11)
Solution method:
(a) Guess values of f1, f2, and f3 (A good starting value is the large Re value for
ε
= 4.5 ×10 −4 or f1 = f2 = f3 = 0.017);
D
(b) Calculate V1, V2, and V3 from Eqs. (9), (10), and (11);
(c) Calculate f1, f2, and f3 from Eqs. (6), (7), (8);
(d) Compare the new fi with the previous ones;
(e) If not good enough agreement, repeat with the new fi as the “guess.”
The results are shown below.
Pipe no.
1
2
3
Re
3.76 E+05
2.19 E+05
1.57 E+05
f
0.0176
0.0184
0.0190
V (m/s)
4.211
2.451
1.760
Q (m3/s)
3.31 E-02
1.93 E-02
1.38 E-02
Problem 8.119
The three water-filled tanks shown in the figure below are connected by pipes as indicated.
If minor losses are neglected, determine the flowrate in each pipe.
Elevation = 60 m
Elevation = 20 m
Elevation = 0
D = 0.10 m
𝓵 = 200 m
f = 0.015
D = 0.08 m
𝓵 = 200 m
f = 0.020
D = 0.08 m
𝓵 = 400 m
f = 0.020
Solution 8.119
Elevation = 60 m
(A)
•
Elevation = 20 m
(B)
•
(1)
Elevation = 0
(C)
•
D = 0.10 m
𝓵 = 200 m
f = 0.015
(2)
(3)
D = 0.08 m
𝓵 = 200 m
f = 0.020
D = 0.08 m
𝓵 = 400 m
f = 0.020
Assume the fluid flows from A to B and A to C. Thus, Q1 = Q2 + Q3 or
π
4
(0.1 m)2V1 =
π
4
(0.08 m)2V2 +
Thus, V1 = 0.64V2 + 0.64V3
π
4
(0.08 m)2V3
(1)
For fluid flowing from A to B with pA = pB = 0 and VA = VB = 0,
 V2
 V2
zA = zB + f1 1 1 + f2 2 2
D1 2 g
D2 2 g
or
 200 m 
 200 m 
V12
V22
60 m − 20 m = (0.015) 
(0.020)
+



 0.1 m  2  9.81 m 
 0.08 m  2  9.81 m 




s2 
s2 


Hence,
40 = 1.529V12 + 2.55V22
Similarly, for fluid flowing from A to C with pA = pC = 0 and VA = VC = 0,
 3 V32
 1 V12
zA = zC + f1
+ f3
D1 2 g
D3 2 g
(2)
or
 200 m 
 400 m 
V32
V12
60 m = (0.015) 
(0.020)
+



 0.1 m  2  9.81 m 
 0.08 m  2  9.81 m 




s2 
s2 


(3)
Hence,
60 = 1.529V12 + 5.10V32
Solve Eqs. (1), (2), and (3) for V1, V2, and V3. From Eqs. (1) and (3):
60 = 1.529(0.64)2 (V2 + V3 )2 + 5.10V32 , or 95.8 = (V2 + V3 )2 + 8.14V32 .
(4)
Subtract Eq. (2) from Eq. (3):
60 − 40 = 5.10V32 + 2.55V22 or V2 = 2V32 − 784
Thus, from Eqs. (4) and (5), 8.14V32 +
This can be simplified to
2V3 2V32 − 7.84 = 103.6 − 11.14V32
(
(5)
2V32 − 784 +V3
) − 95.8 = 0
2
(6)
Square both sides and rearrange to give V34 − 19.63V32 + 92.5 = 0 , which can be solved by the
quadratic formula to give
19.632 − 4(92.5)
m
m
= 11.77 or 7.86. Thus V3 = 3.43
or V3 = 2.80 .
2
s
s
19.63 ±
V32 =
m
is not a solution of the original equations, Eqs. (1), (2), and
s
(3). With this value the right-hand side of Eq. (6) is negative
(i.e., 103.6 − 11.14V32 = 103.6 − 11.14(3.43) 2 = − 24.5 ). As seen from the left-hand side of Eq.
(6), this cannot be. This extra root was introduced by squaring Eq. (6).
Note: The value V3 = 3.43
Thus, Q3 = A3V3 =
π
m
m3

(0.08 m)2  2.80  = 0.0141
4
s 
s

Also, from Eq. (3):
60 = 1.529V12 + 5.10(2.80)2 or
or Q1 = AV
1 1=
V1 = 3.62
m
s
π
m
m3

(0.10 m)2  3.62  = 0.0284
4
s 
s

and from Eq. (1):
3.62 = 0.64V2 + 0.64(2.80) or
V2 = 2.86
m
s
or
Q2 = A2V2 =
π
m
m3

(0.08 m)2  2.86  = 0.0143
4
s 
s

(Note: An equation solver would allow one to program the equations and determine the
flows directly.)
Problem 8.120
Five oil fields, each producing an output of Q barrels per day, are connected to the 28-in.diameter “mainline pipe” (A – B – C) by 16-in.-diameter “lateral pipes” as shown in the
figure below. The friction factor is the same for each of the pipes and elevation effects are
negligible. (a) For section A – B, determine the ratio of the pressure drop per mile in the
mainline pipe to that in the lateral pipes. (b) Repeat the calculations for section B – C.
Q
Lateral
Q
Q
A
Q
B
C
Q
Main line
Solution 8.120
 V2
1

=f
For any of the pipe sections,
or Δp = ρV 2 f
γ
D 2g
2
D


1
1
2
2
(a) Thus, ΔpAB = ρVAB
fAB AB and Δplat = ρVlat
flat lat , where fAB = flat
2
2
DAB
Dlat
Δp
Hence,
2
ΔpAB VAB
 AB
D
= AB
2
Δplat
Vlat
 lat
D
(1)
lat
Also,
D 
V
QAB = 3Q so that D ABVAB = 3 D latVlat or AB = 3  lat 
Vlat
4
4
 DAB 
π
2
π
2
Thus, Eq. (1) becomes
ΔpAB
2
5
5
  D 2   D 
 Dlat 
 AB
 16 in. 
lat
lat


= 3
= 9
 = 9
 = 0.548
Δplat   DAB    DAB 
DAB 
 28 in. 



 lat
(b) Similarly, for section BC:
ΔpBC 1
f
2
so that
= ρVBC
 BC
2
DBC
2
ΔpBC VBC
D
 BC
= BC
2
Δplat
Vlat
 lat
D
lat
(2)
2
Also,
2
D 
V
QBC = 5Qlat, or BC = 5  lat  so that Eq. (2) becomes
Vlat
 DBC 
ΔpBC
2
5
5
  D 2   D 
 Dlat 
 16 in. 
 BC
lat
lat

= 5
= 25 
 = 25 
 = 1.52
Δplat   DBC    DBC 
28 in. 
DBC 




 lat
Problem 8.122
Water flows through the orifice meter shown in the figure below at a rate of 0.10 cfs. If d =
0.1 ft, determine the value of h and the pressure difference associated with this h.
h
d
Q
2 in.
Solution 8.122
Q = C0 A0
2( p1 − p2 )
ρ (1 − β )
4
where β =
d 0.1 ft
=
= 0.6, p1 − p2 = γ h = ρ gh
2
D
ft
12
(1)
Also,
ft 3
Q
s = 4.58 ft so that
=
V=
π 2 π  2 2
s
D
ft


4
4  12 
0.10
ft   2 

4.58   ft 

VD 
s   12 
=
= 6.31× 10 4 . Hence, from the figure below, C0 = 0.616
Re =
2
v
ft
1.21× 10 −5
s
0.66
D
0.64
D
D
2
V
d
C0 0.62
β=
0.60
0.58 4
10
105
106
Re = ρVD/μ
d
= 0.7
D
0.6
0.5
0.4
0.2
107
108
Therefore, from Eq. (1):
ft
π
= (0.616) (0.1ft)2
s
4
3
0.10
ft 

2 ρ  32.2 2  h
s 

ρ (1 − 0.6 4 )
lb 
lb

 p = γ h =  62.4 3  ( 5.77 ft ) = 360 2 = 25 psi
ft 
ft

or h = 5.77 ft
Problem 8.123
Water flows through the orifice meter shown in the figure below such that h = 1.6 ft with d
= 1.5 in. Determine the flowrate.
h
d
Q
2 in.
Solution 8.123
Q = C0 A0
2( p1 − p2 )
ρ (1 − β )
4
where β =
d 1.5 in.
=
= 0.75, and p1 − p2 = γ h = ρ gh
D 2 in.
(1)
Thus,
1
 
ft 
2
2  2 ρ  32.2  (1.6 ft) 
π  1.5 
s

Q = C0 
ft   
4  12  
ρ (1 − 0.754 )



or
Q = 0.151C0
 2 
V  ft 
VD
 12  = 1.38 × 10 4V , where V = Q = 45.8Q
Also, Re =
=
π 2
v
ft 2
D
1.21× 10 −5
4
s
Trial and error solution:
ft3
Assume C0 = 0.6; or from Eq. (1), Q = 0.151(0.6) = 0.0906
.
s
Hence, from Eq. (2), V = 45.8(0.0906) = 4.15 and Re = 5.73 × 104
(2)
0.66
D
0.64
D
D
2
V
d
C0 0.62
β=
0.60
0.58 4
10
105
106
d
= 0.7
D
0.6
0.5
0.4
0.2
107
108
Re = ρVD/μ
From the figure above, with this Re and β, C0 = 0.62 ≠ 0.6 (the assumed value)
ft3
ft
. Thus V = 45.8(0.0936) or V = 4.29 and
Assume C0 = 0.62 or Q = 0.151(0.62) = 0.936
s
s
4
Re = 5.92 × 10 . From the figure above, C0 = 0.62, the assumed value.
ft 3
Hence, Q = 0.0936
s
Problem 8.124
Water flows through the orifice meter shown in the figure below at a rate of 0.10 cfs. If h =
3.8 ft, determine the value of d.
h
d
Q
2 in.
Solution 8.124
Q = C0 A0
2( p1 − p2 )
ρ (1 − β )
4
, where β =
d
d
=
, p1 − p2 = γ h = ρ gh
D 2 ft
12
(1)
and
ft   2 

ft3
4.58   ft 
0.10

Q
s   12 
s = 4.58 ft , Re = VD = 
=
V=
= 6.31× 10 4
2
2
π 2 π2 
s
v
ft
D
1.21× 10 −5
 ft 
4
s
4  12 
0.66
D
0.64
D
D
2
V
d
C0 0.62
β=
0.60
0.58 4
10
105
106
Re = ρVD/μ
d
= 0.7
D
0.6
0.5
0.4
0.2
107
108
Trial and error solution: From Eq. (1)
1
 
2
ft 
2 ρ  32.2 2  (3.8 ft) 
3

C d2
π
ft
s 
 or 0.00814 = 0
(2)
= C0 ( d ft)2  
0.1
, where d ~ ft
s
4
ρ (1 − β 4 )


1− β 4


2
2
Assume β = 0.6, or d = β = 0.6 = 0.10 ft. Thus from Eq. (2), C0 = 0.759. However, from
12
12
the figure above, for this Re and β, C0 = 0.615 ≠ 0.759. Assume β = 0.65 or
2
d = 0.65 = 0.108 ft . From Eq. (2), C0 = 0.633. From the figure above, C0 = 0.618 ≠ 0.633.
12
2
Assume β = 0.67, or d = 0.67 = 0.112 ft. From Eq. (2), C0 = 0.580. From the figure above,
12
C0 = 0.619 ≠ 0.580 Thus, d ≈ 0.109 ft
Problem 8.125
Water flows through a 40-mm-diameter nozzle meter in a 75-mm-diameter pipe at a rate of
m3
0.15
. Determine the pressure difference across the nozzle if the temperature is (a) 10 °C,
s
or (b) 80 °C.
Solution 8.125
V
d = 0.04 m
D = 0.075 m
m3
Q = 0.015 s
Q = Cn An
2( p1 − p2 )
ρ (1 − β )
4
, where β =
d 40 mm
=
= 0.533
D 75 mm
Thus,
0.015
2( p1 − p2 )
π
m3
= C n (0.04 m)2
s
4
ρ (1 − 0.5334 )
or
Cn
1
kg
p1 − p2 = 8.09 ρ 2 , where ρ ~ 3 , p1 − p2 ~
m
Also, Re =
N
(1)
m2
m3
0.015
s
Q
m
VD V (0.075 m)
=
= 3.40
=
, with V =
π 2 π
s
v
v
D
(0.075 m)2
4
4
(a) Assume T = 10 °C, or from Table B.2 Physical Properties of Water (SI Units):
ρ = 999.7
kg
m3
, v = 1.307 ×10
−6 m
2
s
Thus,
m

 3.40  (0.075 m)
s 
= 1.95 ×105 so that from the figure below:
Re = 
2
m
1.307 ×10 −6
s
Cn = 0.986
1.00
0.2
0.4
0.6
0.98
Cn
β=
d
= 0.8
D
0.96
D
0.94
104
105
V
106
d
107
108
Re = ρVD/μ
From Eq. (1): 0.986
1
p1 − p2 = 8.09(999.7) 2 or p1 − p2 = 6.73 × 10 4
N
m2
Thus, p1 – p2 = 67.3 kPa
(b) Assume T = 80 °C, or from the table above: ρ = 971.8
kg
m
, v = 3.65 ×10 −7
3
m

 3.40  (0.075 m)
s 
Thus, Re = 
= 6.99 ×105 so that from the figure above:
2
m
3.65 ×10 −7
s
Cn = 0.991
From Eq. (1): 0.991
1
p1 − p2 = 8.09(971.8) 2 or p1 − p2 = 6.48 × 10 4
Thus, p1 – p2 = 64.8 kPa
N
m2
m2
,
s
Problem 8.126
m3
. Determine the
s
pressure drop across a flow nozzle placed in the line if the nozzle diameter is 20 mm.
Gasoline flows through a 35-mm-diameter pipe at a rate of 0.0032
Solution 8.126
Q = Cn An
2( p1 − p2 )
ρ (1 − β 4 )
, where β =
d 20 mm
π
=
= 0.571, An = d 2
D 35 mm
4
(1)
From Table 1.6 Approximate Physical Properties of Some Common Liquids (SI Units)
kg
N ⋅s
ρ = 680 3 and μ = 3.1 × 10 −4 2
m
m
m3
0.0032
Q
ρVD
s = 3.33 m
=
, where V =
Thus, Re =
π 2 π
s
μ
(0.035 m)2
D
4
4
So that
kg  
m

 680 3   3.33  (0.035 m)
s 
m 
Re = 
= 2.56 × 105
N
s
⋅
3.1× 10 −4 2
m
1.00
0.2
0.4
0.6
0.98
Cn
β=
d
= 0.8
D
0.96
D
0.94
104
105
V
106
d
107
108
Re = ρVD/μ
Hence, from the figure above, Cn = 0.986
From Eq. (1),
0.0032
2( p1 − p2 )
m3
π
= (0.986) (0.020 m)2
kg 
s
4

4
 680 3  1 − 0.571

m 
(
or
p1 − p2 = 3.24 × 10 4
N
m2
= 32.4 kPa
)
Problem 8.127
lb
. Determine the
s
pressure at the 2-in.-diameter throat of a Venturi meter placed in the pipe.
Air at 200 °F and 60 psia flows in a 4-in.-diameter pipe at a rate of 0.52
Solution 8.127
Q = CV AT
2( p1 − p2 )
ρ (1 − β )
4
, where β =
d 2 in.
lb
=
= 0.5 and γ Q = 0.52
D 4 in.
s
(1)
lb  
in.2 

60
144




ft 2 
 in.2  
p
slug
Also, ρ =
=
= 7.63 × 10 −3 3
RT 
ft ⋅ lb 
ft
 1716 slug ⋅°R  (200 + 460)°R


so that


γ = ρ g =  7.63 ×10−3
slug 
ft 
lb
=
32.2
0.246



s
ft3  
ft3
ft3
lb
2.11
0.52
3
s = 24.2 ft
s = 2.11 ft and V = Q =
Thus, Q =
π 2 π  4 2
lb
s
s
D
0.246 3
ft


4
ft
4  12 
Physical Properties of Air at Standard Atmospheric Pressure (BG/EE Units)a
Temperature
(°F)
Density, ρ
(slugs/ft3)b
Specific
Weightc, γ
(lb/ft3)
−40
−20
0
10
20
30
40
50
60
70
80
90
100
120
140
2.939 E – 3
2.805 E – 3
2.683 E – 3
2.626 E – 3
2.571 E – 3
2.519 E – 3
2.469 E – 3
2.420 E – 3
2.373 E – 3
2.329 E – 3
2.286 E – 3
2.244 E – 3
2.204 E – 3
2.128 E – 3
2.057 E – 3
9.456 E – 2
9.026 E – 2
8.633 E – 2
8.449 E – 2
8.273 E – 2
8.104 E – 2
7.942 E – 2
7.786 E – 2
7.636 E – 2
7.492 E – 2
7.353 E – 2
7.219 E – 2
7.090 E – 2
6.846 E – 2
6.617 E – 2
Dynamic
Viscosity,
μ (lb ·
s/ft2)
Kinematic
Viscosity, ν
(ft2/s)
Specific
Heat
Ratio, k
(—)
Speed
of
Sound,
c (ft/s)
3.29 E – 7
3.34 E – 7
3.38 E – 7
3.44 E – 7
3.50 E – 7
3.58 E – 7
3.60 E – 7
3.68 E – 7
3.75 E – 7
3.82 E – 7
3.86 E – 7
3.90 E – 7
3.94 E – 7
4.02 E – 7
4.13 E – 7
1.12 E – 4
1.19 E – 4
1.26 E – 4
1.31 E – 4
1.36 E – 4
1.42 E – 4
1.46 E – 4
1.52 E – 4
1.58 E – 4
1.64 E – 4
1.69 E – 4
1.74 E – 4
1.79 E – 4
1.89 E – 4
2.01 E – 4
1.401
1.401
1.401
1.401
1.401
1.401
1.401
1.401
1.401
1.401
1.400
1.400
1.400
1.400
1.399
1004
1028
1051
1062
1074
1085
1096
1106
1117
1128
1138
1149
1159
1180
1200
160
180
200
300
400
500
750
1000
1500
1.990 E – 3
1.928 E – 3
1.870 E – 3
1.624 E – 3
1.435 E – 3
1.285 E – 3
1.020 E – 3
8.445 E – 4
6.291 E – 4
6.404 E – 2
6.204 E – 2
6.016 E – 2
5.224 E – 2
4.616 E – 2
4.135 E – 2
3.280 E – 2
2.717 E – 2
2.024 E – 2
4.22 E – 7
4.34 E – 7
4.49 E – 7
4.97 E – 7
5.24 E – 7
5.80 E – 7
6.81 E – 7
7.85 E – 7
9.50 E – 7
2.12 E – 4
2.25 E – 4
2.40 E – 4
3.06 E – 4
3.65 E – 4
4.51 E – 4
6.68 E – 4
9.30 E – 4
1.51 E – 3
a
1.399
1.399
1.398
1.394
1.389
1.383
1.367
1.351
1.329
1220
1239
1258
1348
1431
1509
1685
1839
2114
Based on data from R. D. Blevins, Applied Fluid Dynamics Handbook, Van Nostrand Reinhold Co., Inc.,
New York, 1984.
b
To obtain EE units (lbm/ft3) multiply by 32.174.
c
Density and specific weight are related through the equation γ = ρg. For this table g = 32.174 ft/s2.
Also, from the table above, μ = 4.49 × 10 −7
lb ⋅ s
so that
ft 2
slug  
ft   4 

7.63 × 10 −3 3   24.2   ft 

ρVD 
s   12 
ft  
Re =
=
= 1.37 × 105
lb
⋅
s
μ


−7
 4.49 × 10

ft 2 

ft 3
Q
ft
s
Also, Vd =
=
= 96.7
2
π 2 π2 2
s
d
ft


4
4  12 
2.11
slug  
ft   2 

1.63 × 10 −3 3   96.7   ft 

ρV
s   12 
ft  
so Re = d = 
= 2.74 × 105
lb
s
⋅
μ
4.49 × 10 −7 2
ft
From Section 8.6.1,
CV = 1.0054 −
6.88
6.88
= 1.0054 −
= 0.99
5
Red
2.74 ×10
From Eq. (1): 2.11
π2 
ft 3
= (0.99)  ft 
s
4  12 
2
or
lb  1 ft 2 
lb
= 0.253

2
2
ft  144 in. 
in.2
Thus, p2 = (60 − 0.253)psia = 59.75psia
p1 − p2 = 36.4
2( p1 − p2 )

−3 slug 
(1 − 0.54 )
 7.63 × 10
3 
ft 

Problem 8.128
A 2.5-in.-diameter flow nozzle meter is installed in a 3.8-in.-diameter pipe that carries water
at 160°F. If the air–water manometer used to measure the pressure difference across the
meter indicates a reading of 3.1 ft, determine the flowrate.
Solution 8.128
air
h = 3.1 ft
V
(1)
D = 3.8 in. •
• d = 2.5 in.
(2)
2( p1 − p2 )
Q = C n An
ρ (1 − β )
4
, where β =
d 2.5 in.
=
= 0.658
D 3.8 in.
(1)
From Table B.1 Physical Properties of Water (BG/EE Units): ρ = 1.896
μ = 8.32 × 10 −6
lb ⋅ s
ft 2
, so that
slugs   3.8 

1.896 3 V 
ft 

ρVD 
1.2 

ft

=
Re =
lb ⋅ s
μ
8.32 ×10 −6 2
ft
or
ft
Re = 7.22 × 10 4V , where V ~
s
Also, with Q =
π
4
(2)
D2V , Eq. (1) becomes (using p1 – p2 = γh):
1
2
 
ft  
slugs 
(3.1 ft) 
3 
π  3.8 
π  2.5
s
ft






ft  V = C n 
ft 

slugs 
4  12 
4  12   
4
1.896 3  (1 − 0.658 ) 
 

ft 
2
2 2  32.2 2   1.896
 
or
V = 6.78C n
(3)
slugs
ft 3
,
1.00
0.2
0.4
0.6
0.98
Cn
β=
d
= 0.8
D
0.96
D
0.94
104
105
V
106
d
107
108
Re = ρVD/μ
Trial and error solution using the figure above for Cn = Cn (Re, β = 0.658):
ft
Assume Cn = 0.99. From Eq. (3), V = 6.78(0.99) = 6.71
s
ft


From Eq. (2), Re = 7.22 ×104  6.71  = 4.84 ×105, which from the figure above gives Cn =
s

0.99 (checks with assumed value)
2
π 2
π  3.8  
ft 
ft
ft
ft   6.71  = 0.528 .
Thus, V = 6.71 and Q = D V = 
4
4  12  
s
s
s
3
Problem 8.129
A 0.064-m-diameter nozzle meter is installed in a 0.097-m-diameter pipe that carries water
at 60 °C. If the inverted air–water U-tube manometer used to measure the pressure
difference across the meter indicates a reading of 1 m, determine the flowrate.
Solution 8.129
air
h=1m
γ
V
(1)
D = 0.097m •
Q = Cn An
• d = 0.064 m
(2)
2( p1 − p2 )
ρ (1 − β )
4
, where β =
d 0.064 m
=
= 0.660
D 0.097 m
(1)
From Table B.2 Physical Properties of Water (SI Units): ρ = 983.2
μ = 4.665 × 10 −4
N ⋅s
m2
kg
m3
,
, so that
kg 

983.2 3 V (0.097m)

ρVD 
m 
=
Re =
N ⋅s
μ
4.665 × 10 −4 2
m
or
Re = 2.04 × 105V , where V ~
Also, with Q =
π
4
m
s
D2V and p1 − p2 = γ h = ρ gh = 983.2
(2)
kg 
m
N
9.81 2  (1 m) = 9.65 ×103 2 Eq.
3
m 
s 
m
(1) becomes
1

N  2

2  9.65 × 103 2  

π
π

m  
(0.097 m)2V = C n (0.064 m)2 
kg
4
4


 983.2
(1 − 0.660 4 ) 
3
 

m 
or
V = 2.14C n
(3)
1.00
0.2
0.4
0.6
0.98
Cn
β=
d
= 0.8
D
0.96
D
0.94
104
105
V
106
d
107
108
Re = ρVD/μ
Trial and error solution using the figure above for Cn = Cn (Re, β = 0.660):
Assume Cn = 0.99. From Eq. (3), V = 2.14(0.99) = 2.12
m
s
From Eq. (2), Re = 2.04 × 105 (2.12) = 4.32 × 105, which from the figure above gives CD =
0.99 which checks with assumed value.
Thus, V = 2.12
π
π
m
m3
m

and Q = D 2V = (0.097 m)2  2.12  = 0.0157
s
4
4
s 
s

Problem 8.130
A 50-mm-diameter nozzle meter is installed at the end of a 80-mm-diameter pipe through
which air flows. A manometer attached to the static pressure tap just upstream from the
nozzle indicates a pressure of 7.3 mm of water. Determine the flowrate.
Solution 8.130
h = 7.3 mm H2O
V
•
(2) d = 50 mm
D = 80 mm
•
(1)
Q = Cn An
p2 = 0
2( p1 − p2 )
ρ (1 − β )
4
, where β =
d 50 mm
=
= 0.625 and
D 80 mm
N 
N

p1 − p2 = γ H2O h =  9800 3  (7.3 ×10−3 m) = 71.5 2

m 
m
Thus, with An =
Q = Cn
π
4
π
4
d 2,
( 0.050 m )2
N 

2  71.5 2 

m 
kg 

4
1.23 3  (1 − 0.625 )

m 
or
m3
Q = 0.0230Cn. Assume Cn = 0.97 so that Q = 0.0223
s
3
m
0.0223
Q
s = 4.44 m
=
and V =
π 2 π
s
D
(0.08 m)2
4
4
or
m

4.44  (0.08 m)

VD 
s 
=
= 2.43 × 10 4. With the Re and γ, we obtain
Re =
2
v
m
1.46 × 10 −5
s
Cn = 0.963 ≠ 0.97 (the assumed value) (See the figure below)
1.00
0.2
0.4
0.6
0.98
Cn
β=
d
= 0.8
D
0.96
D
0.94
104
105
V
106
d
107
108
Re = ρVD/μ
Thus, assume Cn = 0.963 so that Q = 0.0230(0.963) = 0.0221
m3
s
m3
0.0221
s = 4.40 m
and V =
π
s
(0.08 m)2
4
m

 4.40  (0.08 m)
s 
Check Cn: With Re = 
= 2.41× 10 4, we obtain Cn = 0.963 from the figure
2
m
1.46 ×10 −5
s
above (checks)
m3
Thus, Q = 0.0221
.
s
Problem 8.131
Water flows through the Venturi meter shown in the figure below. The specific gravity of
the manometer fluid is 1.52. Determine the flowrate.
6 in.
Q
3 in.
2 in.
SG = 1.52
Solution 8.131
V
Q
(2)
• 3 in.
(1)
6 in. •
ℓ
γm
h = 2 in.
SG = 1.52
Q = CV AT
2( p1 − p2 )
ρ (1 − β 4 )
, where β =
d 3 in.
=
= 0.5
D 6 in.
Also,
p1 + γ ℓ = p2 + γ (ℓ − h) + γ (SG)h or p1 – p2 = γ (SG − 1)h = ρg(SG – 1)h
Hence,
Q = CV AT
2 ρ g (SG − 1)h
ρ (1 − β 4 )
or
1
 
ft 
 2 2
ft 
2  2  32.2 2  (1.52 − 1) 
π3 
12  

s



Q = CV  ft  
4  12  

(1 − 0.54 )


Thus,
Q = 0.1198CV
Thus Q = 0.1198 CV
Red =
Vd d
ν
where Vd =
Q
π
4
Hence, Red =
Q
π
4
dν
=
d2
Q

−5 ft

 ft   1.21× 10
4  12  
s 
π3
2
=
Q
2.38 × 10 −6
ft 3
s
Assuming 103 < Red < 5 ×105, CV = 1.0054 −
Hence, CV = 1.0054 −
6.89
Q
2.38 × 10 −6
so




6.88


Q = 0.1198 1.0054 −


Q


2.38 ×10 −6 

Using a root solving method, Q = 0.117
ft 3
s
checking Re,
Re =
Q
2.38 × 10 −6
2
ft
s
which is < 5 × 105.
=
0.117
2.38 × 10
−6
= 4.9 × 10 4
6.98
from Section 8.6.1
Red
Problem 8.132
Air flows through the Venturi meter shown in the figure below. The specific gravity of the
manometer fluid is 1.52. Determine the flowrate. Would compressibility effects be
important?